Excel in Mathematics Book 10 Final_CTP (2080)_compressed (1)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 101 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid Step 3: Fill the same cone with water up to the brim for the second time and empty it into the same cylinder. We see that the cylinder is still not complete with water. Step 4: Fill the same cone with water up to the brim for the third time and empty it into the same cylinder. It is observed that the cylinder also fills up to the brim. From the above experiment, we observed that the volume of the cylinder is equal to three times the volume of a cone. Thus, volume of 3 cones = volume of a cylinder = Area of base (A) × Height (h) = πr2 h ∴ Volume of a cone = 1 3 πr2 h Facts to remember In a cone having radius of base r units and height h units; 1. Circumference of base = 2πr 2. Area of base = πr2 3. C.S.A. = πrl 4. T.S.A. = πr (r + l) 5. Volume of cone = 1 3 × Area of base × height= 1 3 πr2 h Worked-out Examples Example 1: From the given right circular cone, calculate it's curved surface area, total surface area and volume. Solution: Here, height of the cone (h) = 4 cm and slant height (l) = 5 cm. Radius of the circular base (r) = 52 – 42 = 25 – 16= 9 = 3 cm Now, curved surface area of the cone = πrl = 22 7 × 3 cm × 5 cm = 47.14 cm2 Also, total surface area of the cone = πr (r + l) = 22 7 × 3 (3 + 5) = 75.43 cm2 And, volume of the cone = 1 3 πr2 h = 1 3 × 22 7 × 3 cm × 3 cm × 4 cm = 37.71 cm3 Example 2: If the total surface area of a cone is 704 cm2 and radius of its base is 7 cm, find the volume of the cone. Solution: Here, the radius of the base of the cone (r) = 7cm, The total surface area of the cone = 704 cm2 5 cm 4 cm

Vedanta Excel in Mathematics - Book 10 102 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid or, πr ( r + l ) = 704 cm2 or, 22 7 × 7 (7 + l) = 704 cm2 or, l = 25 cm Also, the vertical height of the cone (h) = l 2 – r2 = 252 – 72 = 24 cm Now, the volume of the cone = 1 3 πr2 h = 1 3 × 22 7 × 7 × 7 × 24cm3 = 1,232 cm3 So, the required volume of the cone is 1,232 cm3 . Example 3: If the volume of the given cone is 1848 cm3 , and the radius of its base is 14 cm, find its curved surface area. Solution: Here, the radius of the base of the cone ( r ) = 14 cm The volume of the cone = 1848 cm3 or, 1 3 πr2 h = 1848 cm3 or, 1 3 × 22 7 × 14 × 14 × h = 1848 cm3 or, h = 9 cm Also, the slant height of the cone (l) = h2 + r2 = 92 + 142 = 16.64 cm Now, the curved surface area of the cone = πrl = 22 7 × 14 × 16.64 cm2 = 732.16 cm2 So, the required curved surface area of the cone is 732.16 cm2 . Example 4: Mrs. Karki designs a birthday hat with circumference of base 10π inch and vertical height of 12 inch. She attaches a piece of elastic along its diameter. (i) How long is the elastic? (ii) How much paper does she need to make the hat? (iii) How much cubic inch of the air does it hold? Solution: Here, circumference of the circular base = 10π inch or, 2π r = 10π ∴ = 5 inch (i) The length of elastic = diameter = 2r = 2 × 5 inch = 10 inch. (ii) Vertical height of the hat (h) = 12 inch Now, the slant height of the cone (l) = h2 + r2 = 52 + 122 = 13 inch Also, C.S.A. of hat = πrl = 22 7 × 5 × 13 sq. inch = 204.29 sq. inch Hence, 204.29 sq. inch of paper is required to make the hat. 7cm 25cm h 14 cm Q O R P h

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 103 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid (iii) Volume = 1 3 πr2 h = 1 3 × 22 7 × 5 × 5 × 12 cu. inch = 314.29 cu. inch Hence, the cap holds 314.29 cu. inch of air. Example 6: Once a group of people went for a picnic at a hill side. Due to peak season, they did not get a proper accommodation there. The weather was fine so they decided to make a conical tent in a park. They went to a tent house near by the picnic sport and brought the canvas in rent that is enough to make the tent with base radius 11.2 m and height 6.6 m. (i) How many people were there in the group if each person required 8.96 square meter of space on the ground for the accommodation? (ii) If they decided to share the rent of the canvas equally, what amount was paid by each person at Rs 15 per sq. m? Solution: Here, in the conical tent, radius of base (r) = 11.2 m and height (h) = 6.6 m (i) Now, area of circular base = πr2 = 22 7 × 11.2 × 11.2 m2 = 394.24 m2 Area of space on the ground required for each person = 8.96 m2 No. of people = Area of circular base Space required for each person = 394.24 m2 8.96 m2 = 44 Hence, there were 44 people. (ii) Also, we know l 2 = r2 + h2 = (11.2 m)2 + (6.6 m)2 = 169 m2 = (13 m)2 ∴ l = 13 m C.S.A. = πrl = 22 7 × 11.2 × 13 m2 = 457.6 m2 Rent of canvas required to make tent = Area × Rate = 457.6 × Rs 15 = Rs 6,864 ∴ Share of each person for rent of the canvas = Rs 6864 44 = Rs 156 Hence, each person paid Rs 156 as the rent of the canvas for the tent. EXERCISE 5.2 General section 1. a) In the given right cone, the radius of circular base = r, vertical height = h and the slant height = l. Write the formulae to find: (i) circumference of base (ii) area of base (iii) curved surface area (iv) total surface area (v) volume (vi) relation among r, h and l. b) If a is the radius of the circular base and b is the slant height of a cone, write the formula to find its curved surface area. c) If x is the radius of the base and y is the slant height of a cone, write the formula to find its total surface area. d) The radius of the circular base of a cone of height y is x, write the formula to find its volume. h r l P O Q

Vedanta Excel in Mathematics - Book 10 104 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid 2. a) Calculate the curved surface area of the following right circular cones. (i) (ii) (iii) (iv) b) The slant height of a right circular cone with radius 7 cm is 15 cm. Find its curved surface area. c) The radius of the circular base of a cone is 14 cm and its slant height is 20cm, find the total surface area of the cone. d) The slant height of a circular cone with radius 21 cm is 25 cm, find the total surface area. 3. a) Find the volume of the following right cones. (i) (ii) (iii) (iv) b) The vertical height of a cone is 15 cm and the radius of its base is 7 cm, find the volume of the cone. c) If the vertical height of a cone with radius 21 cm is 12 cm, find the volume of the cone. d) A group of student is making a model of a mountain with clay in the shape of right circular cone. If the mountain is 3 feet tall and the radius of the base is 3.5 feet, what is the volume of clay needed to make the mountain? 4. a) The curved surface area of a cone is 220 cm2 . If the diameter of its circular base is 14 cm, find its slant height. b) If the total surface area of a cone is 594 cm2 and its slant height is 20 cm, find the radius of its circular base. c) The volume of a cone is 23.1 cm3 . If the radius of its base is 2.1 cm, find the vertical height of the cone. Creative section-A 5. Find the curved surface area and total surface area of the following cones. a) b) c) d) 7cm 25 cm 10cm 14 cm 28 cm 50 cm 30 cm 21 cm 6cm 7cm 7cm 24cm 35 cm 30 cm 42 cm 40 cm 24cm A O B 7cm 21cm P Q O 20cm M O N 28cm 35cm B O 65cm 56cm

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 105 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid 6. Find the volume of the following right circular cones. a) b) c) d) 7. a) The diameter of circular base of a right cone is 10 cm and its total surface area is 90π cm2 . Find its (i) slant height (ii) vertical height (iii) volume b) The diameter of circular base of a right cone is 60 cm and its volume is 12,000π cm3 . Find its (i) height (ii) slant height (iii) total surface area c) The slant height of a right circular cone is 50 cm and its total surface area is 2,816 cm2 . Find its (i) radius of base (ii) vertical height (iii) volume Creative section-B 8. a) On the occasion of daughter’s birthday, Rahul designs a birthday hat with circumference of base 44 cm and vertical height of 24 cm. He attaches a piece of elastic along its diameter. (i) How long is the elastic? (ii) How much paper does he require to make the hat? (iii) How much cubic centimetres of the air does it hold? b) A Vietnamese leaf hat is in the shape of a right circular cone. The circumference of its circular base is 66 inch and a slant height of 13.7 inch. (i) How long is its radius? (ii) How many sq. inch of material is required to make the hat? (iii) How much cu.inch of the air does it hold? 9. a) Once Sumesh and his friends went for a picnic at a hill side. Due to peak season, they did not get a proper accommodation there. The weather was fine so they decided to make a conical tent in a park. They went to a tent house near by the picnic sport and brought the canvas enough to make the tent with base radius 10.5 m and height 10 m for rent. (i) How many people were there in the group if each person required 7.7 square meter of space on the ground for the accommodation? (ii) If they paid the rent of the canvas equally, how much amount was added to each person’s bill at Rs 30 per sq. m? b) Due to heavy floods in a province of Nepal, thousands of people were victimized last year. Immediately an organization of 1,200 schools collectively offered to the province government to provide the canvas for 2,500 conical tents each of having base radius 5.6 m and height 4.2 m. They decided to share the whole expenditure equally. (i) How much canvas was provided for the tents? (ii) If the canvas used to make the tents cost Rs 450 per square meter, what was the amount shared by each school to set-up the tents? 24cm 25cm P Q O 35cm 37cm A T O A P B O 16cm 17cm 132cm 35cm

Vedanta Excel in Mathematics - Book 10 106 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid Project Work and Activity Section 10. a) Make the groups of your at least 5 friends. Draw two circles of same radii as large as possible on the chart paper and cut them out. Cut three sectors of different sizes and roll to form the cones. Measure the radii of base and slant heights. Then, find the curved surface area of each cone. b) Make a hollow right circular cone and a hollow right-circular cylinder of the same height and base radius such that it is open at the top and closed at the bottom. Fill the cone up to the brim with sand and pour it into the cylinder. Repeat the same process until the cylinder gets completely filled up with the sand. Write a short note on the relationship between the volumes of cone and cylinder. OBJECTIVE QUESTIONS 1. In a square based-pyramid, each side of base is a and slant height is l, then the area of its triangular surfaces is (A) al (B) 2al (C) a2 + 2al (D) a2 – 2l 2. The relationship among the base side a, height h and slant height l in a cone is (A) a2 + h2 = l 2 (B) a2 + h2 = l 2 (C) a2 + h2 = l 2 (D) a2 – h2 = l 2 3. What is the vertical height of a square based pyramid in which the side length of base is 16 cm and slant height is 17 cm? (A) 6 cm (B) 9 cm (C) 12 cm (D) 15 cm 4. A square-based pyramid shaped crystal is 6 cm high and its side length of the base is 7 cm. The volume of the crystal is (A) 133 cm3 (B) 120 cm3 (C) 98 cm3 (D) 84 cm3 5. How much canvas is required to make a square based pyramid shaped tent having base length 18 feet and slant height 15 feet? (A) 270 sq. ft. (B) 540 sq. ft. (C) 864 sq. ft. (D) 1,296 sq. ft. 6. The volume of a right cone having base radius r cm and height h cm is (A) πr2 h (B) 13 πr2 h (C) πrh2 (D) 13 πrh2 7. The formula to calculate the total surface area of right cone is (A) πrl (B) 2πrl (C) πr (r + l) (D) 2πr (r + l) 8. The diameter of a cone with height 24 cm and slant height 25 cm is (A) 7 cm (B) 14 cm (C) 17 cm (D) 1 cm 9. How much paper is required to make a conical hat of base radius 14 cm and slant height 11 cm? (A) 484 cm2 (B) 968cm2 (C) 1,100 cm2 (D) 3,388 cm2 10. A cone with base radius is 12 cm has 21 cm height. Its volume is (A) 3,168 cm3 (B) 5,544 cm3 (C) 12 cm3 (D) 12 cm3

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 107 Vedanta Excel in Mathematics - Book 10 Unit 6 Mensuration (II): Combined Solids 6.1 Combined solids –introduction Let’s observe the shapes of the following objects and discuss on the questions given below. Tent Ice-cream House with pyramid roof Capsule (a) What are the basic shapes combined to form a circus tent? (b) What is the shapes of a cone filled with ice-cream? (c) Name the shapes consisting by the house? (d) Which are the shapes combined in the capsule? In our day-to-day lives, we see many shapes that are a blend of different basic shapes. A combination of a solid is that figure which is formed by combining two or more different solids. 6.2 Surface area and volume of some basic 3-D shapes – Review S.N. Name of solid Figure L.S.A. or C.S.A. T.S.A. Volume 1. Square-based prism a a h L.S.A. = 4ah 2a ( a + 2h) a2 h 2. Square-based pyramid a h l L.S.A. =2al a2 + 2al 1 3 a2 h

Vedanta Excel in Mathematics - Book 10 108 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids 3. Cylinder r h C.S.A. = 2πrh 2πr (r + h) πr2 h 4. Sphere r C.S.A. = 4πr2 4πr2 4 3 πr3 5. Hemi-sphere r C.S.A. = 2πr2 3πr2 2 3 πr3 6. Right circular cone h r l C.S.A. = πrl πr (r + l) 1 3 πr2 h 6.3 Combination of square-based prism and pyramid In the figure, a square-based pyramid having base side a, height h and slant height l, and a prism (cuboid) having base side a, height h1 are combined together. Here, L.S.A. of prism = 2h1 (l + b) = 2h1 (a + a) = 4ah1 (a) L.S.A. of combined solid = L.S.A. of pyramid + L.S.A. of prism = 2al + 4ah1 (b) T.S.A. of combined solid = Area of visible and touchable faces = L.S.A. of pyramid + L.S.A. of prism + Area of base = 2al + 4ah1 + a2 (c) Volume of combined solid = Volume of pyramid + Volume of prism = 1 3 a2 h + a2 h1 = a2 (h 3 + h1 ) Facts to remember If a pillar is a combined structure of a square-based pyramid and prism (cuboid) and fixed on the ground, then the surface area of the pyramid = L.S.A. of pyramid + L.S.A. of prism. h l a a a h1

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 109 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids 6.4 Combination of square-based pyramids In the given figure, a square-based pyramid having side a, height h1 and slant height l 1 , and another square-based pyramid having side a, height h2 and slant height l 2 are combined together. Now, (a) L.S.A. of combined solid = L.S.A. of uppermost pyramid + L.S.A. of lowermost pyramid = 2al1 + 2al2 = 2a (l 1 + l 2 ) (b) T.S.A. of combined solid = Area of visible and touchable faces = L.S.A. of uppermost pyramid + L.S.A. of lowermost pyramid = 2al1 + 2al2 = 2a (l 1 + l 2 ) (c) Volume of combined solid = Volume of uppermost pyramid + Volume of lowermost pyramid = 1 3 a2 h1 + 1 3 a2 h2 = 1 3a2 (h1 + h2 ) Worked-out Examples Example 1: The solid given alongside is a right square-based prism with a right pyramid on its top. Find: (i) the volume of uppermost pyramid (ii) the volume of lowermost prism (cuboid) (iii) the volume of the combined solid. Solution: Here, the solid figure is the combination of a square-based pyramid and a prism. (i) For the uppermost square-based pyramid: side length of the base (a) = 12 cm vertical height (h) = 25 cm – 17 cm = 8 cm Now, the volume of the pyramid = 1 3 a2 h = 1 3 × 12 × 12 × 8 cm3 = 384 cm3 (ii) For the lowermost square-based prism; side length of the base (a) = 12 cm vertical height (h1 ) = 17 cm h1 = h2 a h1 a l 1 h2 a l 2 l 2 l 1 12 cm 12 cm 25 cm 17 cm

Vedanta Excel in Mathematics - Book 10 110 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids Also, the volume of the prism = a2 h1 = 12 × 12 × 17 cm3 = 2448 cm3 (iii) The volume of combined solid = volume of pyramid + volume of prism = 384 cm3 + 2448 cm3 = 2,832 cm3 Hence, the volume of uppermost pyramid, lowermost prism and combined solid are 384 cm3 , 2,448cm3 and 2,832 cm3 respectively. Example 2: The solid given alongside is composed up of a right pyramid and a prism having common base. Find: (i) the lateral surface area of uppermost pyramid (ii) the lateral surface area of lowermost prism (cuboid) (iii) the total surface area of the given combined solid. Solution: Here, the solid figure is the combination of a square-based pyramid and a prism. (i) For the uppermost square-based pyramid; side length of the base (a) = 2.5 m vertical height (h) = 10 m – 7 m = 3 m Now, slant height, OQ (l) = OP2 + PQ2 = 32 + 1.252 = 3.25 m ∴The lateral surface area (L.S.A.) = 2al = 2 × 2.5 m × 3.25 m = 16.25 m2 (ii) For the lowermost square-based prism; side length of the base (a) = 2.5 m vertical height (h1 ) = 7 m ∴The lateral surface area = 4ah1 = 4 × 2.5 m × 7 m = 70 m2 (iii) Also, area of base = a2 = (2.5 m)2 = 6.25 m2 Again, The total surface area of combined solid = L.S.A. of pyramid + L.S.A. of prism + area of base = 16.25 m2 + 70 m2 + 6.25 m2 = 92.5 m2 Hence, the lateral surface area of uppermost pyramid and lowermost prism are 16.25 m2 and 70 m2 respectively and the total surface area of combined solid is 92.5 m2 . 2.5m 2.5m 7m 10m 2.5m P Q O 3m 1.25m

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 111 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids Example 3: The adjoining figure is the solid crystal formed by two pyramids of common square base and equal heights. Find the volume and total surface area of the crystal. Solution: Here, the side of square-base of each of the pyramid (a) = 16 cm Let, h1 and h2 be the heights of uppermost and lowermost pyramids respectively. Then, the total height =h1 + h2 = 12 cm Now, Volume of crystal = volume of uppermost pyramid + volume of lowermost pyramid = 1 3 a2 h1 +1 3 a2 h2 = 1 3 a2 (h1 + h2 ) = 1 3 × 162 × 12 cm3 = 1024 cm3 Also, in a square-based pyramid, slant height (l) = a 2 2 + h2 = 16 2 2 + 62 = 82 + 62 = 10 cm Here, the uppermost and lowermost pyramids have equal base and heights. The slant height of uppermost pyramid (l 1 ) = the slant height of lowermost pyramid (l 2 ) = 10 cm ∴The total surface area of the crystal =L.S.A. of uppermost pyramid + L.S.A. of lower most pyramid = 2al1 + 2al2 =2a(l 1 +l 2 ) = 2 × 16 (10 cm + 10 cm) cm2 = 640 cm2 Hence, the volume of the crystal is 1024 cm3 and its total surface area is 640 cm2 . Example 4: The volume of the solid given alongside is 448 cm3 . Find the height of the prism. Solution: Here, the length of the base of uppermost pyramid (a) = 8 cm The slant height of the pyramid (l) = 5 cm The length (l 1 ) = the breadth (b1 ) of cuboid (prism) = 8 cm In the uppermost pyramid, its height (h) = AO = AB2 – OB2 = 52 – 42 = 3 cm Now, the volume of the pyramid = 1 3 a2 × h = 1 3 × 8 × 8 × 3cm3 = 64 cm3 A B 8cm 5cm 8cm O 12 cm 16 cm

Vedanta Excel in Mathematics - Book 10 112 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids The volume of the cuboid (prism) = volume of solid – volume of pyramid = 448 cm3 – 64 cm3 = 384 cm3 Again, the volume of the prism = l 1 × b1 × h1 or, 384 = 8 × 8 × h1 or, h1 = 6 cm Hence, the required height of the prism is 6 cm. Example 5: The gate of a stadium has two pillars, each of height is 8 ft. with four visible lateral faces and 6 ft. × 6 ft. bases. The top of each pillar has a combined pyramid of height 4 feet. If the combined structures of both pillars and pyramids are painted at the rate of Rs 75 per sq. ft., calculate the total cost of painting. Solution: Here, for the uppermost pyramid part of a combined structure: Length of the square base (a) = 6 ft. Vertical height (h) = 4 ft. Now, slant height (l) = a 2 2 + h2 = 6 2 2 + 42 = 42 + 32 ft. = 5 ft. ∴ The lateral surface area (L.S.A.) = 2al = 2 × 6 ft. × 5 ft. = 60 sq. ft. For the lowermost square-based pillar (cuboid) of the combined structure: side length of the base (a) = 6 ft. vertical height (h1 ) = 8 ft. ∴The lateral surface area = 2h1 (l + b) = 2h1 (a + a) = 4ah1 = 4 × 6 ft. × 8 ft. = 192 sq. ft. Also, The total surface area of a combined structure = L.S.A. of pyramid + L.S.A. of prism = 60 sq. ft. + 192 sq. ft. = 252 sq. ft. ∴The total area of two combined structures = 2 × 252 sq. ft. = 504 sq. ft. Again, the rate of cost of painting = Rs 75 per sq. ft. ∴ The total cost of painting = Area × Rate = 504 × Rs 75 = Rs 37,800 Hence, the required cost of painting the pillars with pyramids is Rs 37,800. 4 ft. 6 ft. 3 ft. 8 ft. 6 ft. 4 ft. 6 ft. 3 ft. 8 ft. 6 ft.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 113 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids EXERCISE 6.1 General section 1. a) The adjoining figure is a solid composed up of a square-based pyramid and a prism (cuboid). Write down the formula to find the following measurements. (i) Area of base (ii) Perimeter of base (iii) L.S.A. of pyramid (iv) L.S.A. of prism (v) L.S.A. of the combined solid (vi) T.S.A. of the combined solid (vii) Volume of pyramid (viii) Volume of prism (ix) Volume of the combined solid b) From the given solid made up of two pyramids with common square base and heights h1 and h2 , write the formulae to find: (i) L.S.A. of uppermost pyramid (ii) L.S.A. of lowermost pyramid (iii) T.S.A. of the solid (iv) Volume of uppermost pyramid (v) Volume of lowermost pyramid (vi) volume of the combined solid Creative section-A 2. Each of the following solid is composed up of a square-based pyramid and a prism (cuboid) having common square base. Find: (i) the volume of pyramid (ii) the volume of prism (iii) the volume of the combined solid. a) b) c) 3. a) The adjoining figure is a solid crystal formed by two pyramids with a common square base. If the heights of both of the pyramids are the same, find the volume of the crystal. b) The figure given alongside is a solid crystal formed with two pyramids having a common square base of side 8 cm. If the height of the crystal is 15 cm, find the volume of the crystal. h1 h2 l a a h1 h2 l 2 l 1 a 12cm 12cm 15cm6cm 9cm 9cm 16cm 10cm 16 cm 22cm 12 cm 3cm 3cm 15 cm 8 cm 8cm

Vedanta Excel in Mathematics - Book 10 114 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids 4. a) The adjoining figure is a solid crystal formed by two pyramids with a common square base of side 12 cm. Find the total surface area of the crystal. b) A solid object is formed by two pyramids of common square base. If the side of square base is 7 cm and the slant heights of the pyramids are 6 cm and 5 cm, what is the total surface area of the solid? Creative section-B 5. The following combined solids are made up of the pyramid and the prism (cuboid) having common base. Find: (i) the lateral surface area of uppermost pyramid. (ii) the lateral surface area of lowermost prism (cuboid). (iii) the total surface area of the given combined solid. a) b) c) 6. a) The adjoining solid object is formed with the combination of a pyramid and a square-based prism. If the volume of the solid object is 900 cu. cm, find the height of the prism. b) In the adjoining solid, a square-based pyramid is situated on the top of a square-based cuboid. If the volume of the solid is 4536 cm3 , find the total surface area of the solid. c) If the total surface area of the combined solid made up of the pyramid and square-based prism in the adjoining figure is 240 cm2 , find the volume of the solid. 6cm 6cm 14cm 10cm 12cm 12cm 18cm 8cm 10cm 10cm 22cm 13cm 18 cm 18cm 22 cm 13cm 10 cm 12cm 11cm 10cm 12cm 6 cm 6cm 4 cm

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 115 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids d) The solid given alongside is the combination of a square-based pyramid and a square-based prism. If the total surface area of the solid is 864 cm2 , find the volume of the solid. 7. a) The given figure is a pillar of height 5 m and 60 cm × 60 cm bases. The top of the pillar has a combined pyramid of height 40 cm. Calculate the cost of painting the structure at the rate of Rs 200 per sq. m. b) The gate of a stadium has two pillars, each of height 10 ft. with four visible lateral faces and 3 ft. × 3 ft. bases. The top of each pillar has combined pyramid of height 2 ft. If the combined structures of both pillars and pyramid are painted at the rate of Rs 80 per sq. ft., calculate the total cost of painting. 6.5 Combination of a cylinder and a hemisphere Here, a cylinder has radius r and height h. Also, a hemisphere has radius r. When the hemisphere is surmounted on the top of the cylinder then we will have the following relations: a) C.S.A. of combined solid = C.S.A. of hemisphere + C.S.A. of cylinder = 2πr2 +2πrh = 2πr (r + h) b) T.S.A. of combined solid = Area of visible and touchable surfaces = C.S.A. of ( hemisphere + cylinder) + Area of base = 2πr2 +2πrh + πr2 = 3πr2 +2πrh c) Volume of combined solid = Volume of cylinder + Volume of hemisphere = πr2 h + 2 3 πr3 = πr2 (h + 2r 3 ) 6.6 Combination of a cylinder and a cone Here, a cylinder has radius r and height h1 . Also, a cone has radius r, height h2 and slant height l. When the cone is placed on the top of the cylinder then we will have the following relations in the combined solid so formed. r h h r r h1 r h2 r h1 r h2 l l 12 cm 12cm 10 cm 40cm 60 cm 60cm 5 m

Vedanta Excel in Mathematics - Book 10 116 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids a) C.S.A. of combined solid = C.S.A. of cone + C.S.A. of cylinder = πrl + 2πrh1 b) T.S.A. of combined solid = Area of visible and touchable surfaces = C.S.A. of ( cone + cylinder) + Area of base = πrl + 2πrh1 + πr2 c) Volume of combined solid = Volume of cylinder + Volume of cone = πr2 h1 + 1 3 πr2 h2 = πr2 (h1 + h2 3 ) 6.7 Combination of cone and hemisphere When a cone having base radius r, height h and slant height l is combined with a hemisphere having radius r, then, the relations in the combined solid so formed are: a) C.S.A. / T.S.A. of combined solid = C.S.A. of cone + C.S.A. of hemisphere = πrl + 2πr2 b) Volume of combined solid = Volume of cone + Volume of hemisphere = 1 3 πr2 h +2 3 πr3 = 1 3 πr2 (h + 2r) 6.8 Combination of two cones When a cone having base radius r, height h1 and slant height l 1 , is combuned with another cone having base radius r, height h2 and slant height l 2 , then, we can derive the following relations. a) C.S.A. / T.S.A. of combined solid = C.S.A. of uppermost cone + C.S.A. of lowermost cone = πrl1 + πrl2 = πr (l 1 + l2 ) b) Volume of combined solid = Volume of uppermost cone + Volume of lowermost cone = 1 3 πr2 h1 + 1 3 πr2 h2 = 1 3 πr2 (h1 + h2 ) www.geogebra.org/classroom/ztypgygr Classroom code: ZTYP GYGR Vedanta ICT Corner Please! Scan this QR code or browse the link given below: r h r l l h r h1 l 1 r h2 l 2 r h1 l 1 r h2 l 2

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 117 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids Worked-out Examples Example 1: The given solid object is made up of a cylinder and a hemisphere having same radius. If the total length of combined solid is 44 cm and the height of cylindrical part is 30 cm. Calculate: (i) the total surface area (ii) the volume of the solid. Solution: Here, height of the combined solid = 44 cm and height of the cylindrical part (h) = 30 cm ∴The common radius (r) = height of hemispherical part = 44 cm – 30 cm = 14 cm Now, (i) The total surface area (T.S.A.) of the solid = C.S.A. of hemisphere + C.S.A. of cylinder + Area of base = 2πr2 + 2πrh + πr2 = 3πr2 + 2πrh = πr(3r + 2h) = 22 7 × 14 (3 × 14 + 2 × 30) cm2 = 44 × 102 cm2 = 4,488 cm2 (ii) The volume of the solid = volume of cylinder + volume of hemisphere = πr2 h+ 2 3 πr3 = πr2 (h + 2r 3 ) = 22 7 × 14 × 14 (30 + 2 × 14 3 ) cm3 = 616 × 39.33 cm3 = 24,227.28 cm3 Hence, the required total surface area of combined solid is 4,488 cm2 and its volume is 24,227.28 cm3 . Example 2: A student made a model of pencil-shaped solid object as shown alongside, find its: (i) radius of base (ii) the total surface area and (iii) the volume. Solution: Here, the total length of the solid = 84 cm and the length of cylindrical shape (h1 ) = 60 cm ∴The height of the conical part (h2 ) = 84 cm – 60 cm = 24 cm Now, (i) In the right angled ∆POQ, OR = PQ2 – OP2 = 252 – 242 = 49 = 7 cm ∴The radius of the cone (r) is 7 cm 30 cm 44 cm 25cm 60 cm 84 cm O24 cm 25 cm R P

Vedanta Excel in Mathematics - Book 10 118 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids (ii) The total surface area of the solid (T.S.A.) = C.S.A. of (cone + cylinder) + Area of base = πrl + 2πrh1 + πr2 = πr (l + 2h1 + r) = 22 7 × 7 (25 + 2 × 60 + 7) cm2 = 22 × 152 cm2 = 3,344 cm2 (iii) The volume of the solid = volume of cylinder + volume of cone = πr2 h1 + 1 3 πr2 h2 = πr2 (h1 + h2 3 ) = 22 7 × 7 × 7(60 + 24 3 ) = 154 × 68 = 10,472 cm3 Hence, the radius of solid is 7cm, total surface area is 3,344 cm2 and its volume is 10,472 cm3 . Example 3: Mr. Karki bought a toy which is in the form of cone mounted on a hemisphere of diameter 7 cm. If the total height of the toy is 15.5 cm, find: (i) the volume of toy (ii) surface area of the toy. Solution: Here, the total height of the toy = 15.5cm The common diameter = 7 cm ∴the common radius (r) = 3.5 cm ∴The height of the conical part (h) = 15.5 cm – 3.5 cm = 12 cm Now, (i) The volume of the toy = volume of cone + volume of hemisphere = 1 3 πr2 h+ 2 3 πr3 = 1 3 πr2 (h + 2r) = 1 3 × 22 7 × 3.5 × 3.5 (12 + 2 × 3.5) cm3 = 12.83 × 19 cm3 = 243.83 cm3 Also, we know, slant height of cone (l) = r2 + h2 = 3.52 + 122 = 156.25 = 12.5 cm (ii) The surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere = πrl + 2πr2 = πr (l + 2r) = 22 7 × 3.5 (12.5 + 2 × 3.5) cm2 = 11 19.5 cm2 = 214.5 cm2 Hence, the volume of the toy is 243.83 cm3 and its surface area is 214.5 cm2 . 15.5 cm 7 cm

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 119 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids Example 4: The given figure is the combined solid made up of a cylinder and a hemisphere having same radius. If the total height of the solid is 23 cm and the total surface area is 1,166 cm2 , find the common radius and height of the cylinder. Solution: Here, height of the combined solid = 23 cm Let, the common radius = r cm. Then, height of hemispherical part = radius of base = r cm ∴The height of the cylindrical part (h) = (23 – r) cm. Now, The total surface area of the solid = C.S.A. of hemisphere+C.S.A. of cylinder + Area of base or, 1166 cm2 = 2πr2 + 2πrh + πr2 or, 1166 = πr(3r + 2h) or, 1166 = πr [3r + 2 (23 – r)] or, 1166 = 22 7 × r(r + 46) or, 371 = r2 + 46r or, r2 + 46r – 371 = 0 or, r2 + 53r –7r – 371 = 0 or, r (r + 53) – 7 (r + 53) = 0 or, (r + 53) (r – 7) = 0 Either, r + 53 = 0 ∴r = – 53 which is not possible as the length radius can never be negative. Or, r – 7 = 0 ∴r = 7 Again, h = 23 – r = 23 cm – 7 cm = 16 cm Hence, the required common radius is 7 cm and the height of the cylinder is 16 cm. Example 5: A cone is placed on a cylinder so that the base of the cone covers exactly the base of the cylinder as shown in the figure. The area of the base of the cylinder is 125 cm2 and the height of the cylinder is 3 cm. If the volume of the whole solid is 625 cm3 , find the height of the solid. Solution: Let, h1 be the height of the cylinder and h2 be the height of the cone. Here, area of the base of the cylinder = πr2 = 125 cm2 Also, height of the cylinder (h1 ) = 3 cm Now, volume of the solid = 625 cm3 or, πr2 h1 + 1 3 πr2 h2 = 625 cm3 or, πr2 (h1 + h2 3 ) = 625 cm3 or, 125 cm2 (3 + h2 3 ) = 625 cm3 or, 3 + h2 3 = 5 cm 23 cm 3 cm

Vedanta Excel in Mathematics - Book 10 120 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids or, h2 3 = 2 cm ∴ h2 = 6 cm Hence, the height of the solid = h1 + h2 = 3 cm + 6 cm = 9 cm. Example 6: A shopkeeper sells the water tanks made up of plastic materials and formed with the combination of a cylinder and a hemisphere. For the use of own house, a person bought two water tanks of the same size, each having the base radius 1.05 m and height 3.5 m from the shop. (i) What is the height of the cylindrical part of each tank? (ii) What is the volume of each tank? (iii) If both the tanks are filled with water at the rate of 40 paisa per litre, what is the total cost of water? Solution: Here, total height of each tank = 3.5 m (i) The radius of base of each tank (r) = height of hemispherical part = 1.05 m ∴The height of cylindrical part (h) = 3.5 m – 1.05 m = 2.45 m (ii) The volume of each tank = volume of cylindrical part + volume of hemispherical part = πr2 h+ 2 3 πr3 = πr2 (h + 2r 3 ) = 22 7 × 1.05 × 1.05 (2.45 + 2 × 1.05 3 ) m3 = 3.465 × 3.15 m3 = 10.91475 m3 (iii) The volume of 2 tanks = 2 × 10.91475 m3 = 21.8295 m3 We know, 1 m3 = 1000 litre ∴ 21.8295 m3 = 21.8295 × 1000 litres = 21,829.5 litres Now, the cost of filling water = 21,829.5 × 40 paisa = 873180 paisa = Rs. 8,731.80 Hence, the cost of filling both the tanks with water is Rs. 8,731.80 EXERCISE 6.2 General section 1. a) The given solid is formed by a cylinder and a hemisphere having equal radii. Write down the formula to find: (i) the area of base (ii) C.S.A. of cylindrical part (iii) C.S.A. of hemispherical part (iv) C.S.A. of combined solid (v) T.S.A. of the combined solid (vi) volume of cylindrical part (vii) volume of hemispherical part (viii) volume of the combined solid 3.5 m 1.05 m 3.5 m 1.05 m r h

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 121 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids b) The solid figure given alongside is made up of a cylinder and a cone. Write down the formula to find: (i) the area of base (ii) C.S.A. of cylindrical part (iii) C.S.A. of conical part (iv) C.S.A. of combined solid (v) T.S.A. of the combined solid (vi) volume of cylindrical part (vii)volume of conical part (viii) volume of the combined solid c) In the given combined solid, a right cone is placed on a hemisphere of equal radii. Write down the formula to find: (i) C.S.A. of hemispherical portion (ii) C.S.A. of conical portion (iii) surface area of combined solid (iv) volume of conical portion (v) volume of hemispherical portion (vi) volume of the combined solid Creative section-A 2. Find the volume of the following combined solids. a) b) c) d) e) f) g) h) i) j) k) l) h1 r h2 l l h r 54 cm 40 cm 14 cm 13 cm 10.5 cm 40 cm 20cm 5cm 8cm 54cm 25cm 30cm 12cm 20cm 10cm 7cm 25cm A O P 8cm 10cm 35cm 49cm 9cm 12cm 10cm 15cm 24cm 13cm 14cm 45cm

Vedanta Excel in Mathematics - Book 10 122 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids 3. Calculate the curved surface area of each of the following combined solid. a) b) c) d) 4. a) A tent is cylindrical up to a height of 9 ft. and conical above it. The diameter of the base is 16 ft. and the total height of the tent is 24 ft. (i) What is the height of the conical part? (ii) What is the slant height of the conical part? (iii) How much canvas is required in making the tent? b) A circus tent is in the shape of a cylinder surmounted by a conical roof. The common diameter is 36 m, the height of cylindrical portion is 6.5 m and the height of the roof from the ground is 40 m. (i) What is the height of the conical part? (ii) What is the slant height of the conical part? (iii) How much canvas is required in making the tent? 5. Find the total surface area of the following combined solids. a) b) c) d) e) f) g) h) 6. a) The given solid is made up of a cylinder and cone having equal radius. If the volume of the given solid is 5,544 cm3 , find: (i) the radius of base (ii) curved surface area of cylindrical part. 5.6 cm 4.2 cm 16 cm 30 cm 8cm 6cm 12cm 20 cm 25 cm 44 cm 9 ft. 24 ft. 16 ft. 6.5 m 40 m 36 m 42cm 18cm 28cm 18cm 12cm 10cm 94 cm 80 cm 31 cm 28 cm 88cm 80cm 12cm 24cm 31cm 15.5cm 3.5cm 14cm 48cm 52.8cm 52 cm 28 cm

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 123 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids b) Given figure is a solid composed up of a cylinder with cone at one end. If the volume of the given solid is 18,018 cm3 , find: (i) the radius of base (ii) volume of conical part. 7. a) Given figure is a solid composed of a cylinder with hemisphere at one end. If the total surface area and the height of the solid are 770 sq.cm and 14 cm respectively, find: (i) the common radius (ii) the volume of hemisphere. b) A solid figure is made up of a cylinder surmounted by a hemisphere of same radius. If the total height of the solid is 21.5 cm and it total surface area is 1100 cm2 , find: (i) the height of the cylinder (ii) the volume of the cylindrical part. 8. a) In the adjoining solid, the area of the circular base is 100 cm2 and the height of the cylinder is 3 cm. If the volume of the whole solid is 500 cm3 , find the height of the solid. b) In the given figure, the base area of the cylinder is 64 cm2 and the height of the cylinder is 4 cm. If the volume of the whole solid is 320 cm3 , find the height of the solid. Creative section-B 9. a) A shopkeeper sells the water tanks made up of plastic materials and formed with the combination of a cylinder and a hemisphere. For the use of own house, a person bought two water tanks of same size from the shop, each having the base radius 1.47 m and height 4 m. (i) What is the height of the cylindrical part of each tank? (ii) What is the volume of each tank? (iii) If both the tanks are filled with water at the rate of 50 paisa per litre, what is the total cost of water? b) A water tank is made up of the combination of a cylinder and a hemisphere. The height of the tank is 4 m and the area of the base is 1.54 m2 . (i) Find radius of the base of the tank. (ii) Find the volume of the tank. (iii) Find the cost of filling the tank with water at 40 paisa per litre. 10. a) A gate has two cylindrical pillars with a hemispherical end on the top of each pillar. The height of each pillar is 9.96 m and the height of each cylinder is 9.75 m. 76 cm 40 cm 14 cm 3cm 4cm

Vedanta Excel in Mathematics - Book 10 124 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids (i) Find the area of visible parts of the pillars. (ii) Find the cost of colouring the surface of both pillars at Rs 500 per sq. m. b) The gate of a stadium has two cylindrical pillars, each of height 5 m and the circumference of each base is 22 m. The top of each pillar has a combined cone of height 1.2 m. If the combined structures of both pillars and cones are painted at the rate of Rs 100 per sq. m, find the total cost of painting the pillars. Project Work and Activity Section 11. a) Take a pencil and sharp it. Measure the radius of base, the length of cylindrical part and the slant height of conical part. Calculate its curved surface area, total surface area and volume. b) Measure the circumference of the circular base of water tank and its total height. Calculate the approximate capacity of the tank. c) Measure the diameter and total height of a cone filled with ice-cream. Then, find the approximate volume of ice-cream. 6.9 Mensuration in household activities The figure given alongside represents a kitchen garden of a family. Mother wants to fence the garden to protect vegetables grown inside it. She discuss with her daughter, who is studying in grade 10, about the cost estimation for fencing it. She asked her daughter to estimate the cost for (i) 2 round fencing, (ii) 4 round fencing, and (iii) 5 round fencing. Here is the daughter's estimation: The perimeter of the garden = AB + BC + CD + DA = 10 m + 15 m + 12 m + 8 m = 45 m She found the cost of per metre of fencing wire in the market is Rs 60. She also estimated the labour charge for this activities. The table given below shows her cost estimation. No. of rounds Length of fencing wire Cost of fencing wire Labour cost Total cost 2 2 × 45 m = 90 m 90 × Rs 60 = Rs 5,400 Rs 2,600 Rs 8,000 4 4 × 45 m = 180 m 180 × Rs 60 = Rs 10,800 Rs 3,200 Rs 14,000 5 5 × 45 m = 225 m 225 × Rs 60 = Rs 13,500 Rs 3,500 Rs 17,000 The daughter recommended for 4 rounds of fencing and the mother approved her idea. Similarly, can you please help your family for the following household activities. • Cost of estimation of your kitchen garden A B C D 12 m 10 m 8 m 15 m

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 125 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids • Cost of estimation of carpeting your rooms of your house • Cost of estimation of plastering the walls of the rooms of your house • Cost of estimation of colouring the walls the room of your house • Cost of estimation of building compound walls in your house • Cost of estimation of building an underground water tank Worked-out Examples Example 1: The figure alongside is a compound of a house. Estimate the cost of fencing the compound with 5 rounds of fencing wire. Rate of cost of per metre of wire = Rs 50 Labourer wages = Rs 800 per labour per day The fencing work requires 2 days for 2 labours. Solution: Here, the perimeter of the compound = AB + BC + CD + DA = 10 m + 17 m + 13 m + 8 m = 48 m The length of wire required for 5 rounds = 5 × 48 m = 240 m The cost of the wire = 240 × Rs 50 = Rs 12,000 The wages of 2 labourers for 1 day = 2 × Rs 800 = Rs 1,600 The wages of 2 labourers for 2 days = 2 × Rs 1,600 = Rs 3,200 ∴ The total cost of estimation of fencing the compound = Rs 12,000 + Rs 3,200 = Rs 15,200 Example 2: The dimension of a rectangular room is 7 m long, 6 m broad and 5 m high. (i) Estimate the cost of carpeting its floor at Rs 110 per sq. m. with Rs 1,200 labourer charge. (ii) Estimate the cost of colouring its walls and ceiling at Rs 150 per sq. m. Solution: Here, the length of the room (l) = 7 m the breadth of the room (b) = 6 m the height of the room (h) = 5 m (i) The area of the floor = l × b = 7 m × 6 m = 42 m2 the cost of the carpet = Area of floor × Rate = 42 m2 × Rs 110 per m2 = Rs 4,620 The cost of carpeting the floor with labourer charge = Rs 4,620 + Rs 1,200 = Rs 5,820 (ii) The area of the 4 walls of the room = 2h(l + b) = 2 × 5 (7 + 6) = 10 × 13 = 130 m2 The area of the ceiling = l × b = 7 m × 6 m = 42 m2 The total area of 4 walls and ceiling = 130 m2 + 42 m2 = 172 m2 The cost of colouring walls and ceiling = Area × Rate = 172 m2 × Rs 150 per m2 = Rs 25,800 A B C D 13 m 10 m 8 m 17 m

Vedanta Excel in Mathematics - Book 10 126 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids Example 3: The figure alongside is a rectangular compound wall of a house. The dimensions of the wall is 20 m long, 15 m broad, 4 m high and 10 cm thick. Estimate the cost of building the walls by using bricks of dimensions 20 cm × 10 cm × 5 cm and rate of cost of bricks is Rs 18,000 per 1000 bricks. Labourer wages is Rs 650 per labourer per day and 2 labourers for 5 days. Solution: Here, the outer length of the wall (l) = 20 m the inner breadth of the wall = 15 m ∴ the thickness of the wall (d) = 10 cm = 0.1 m ∴ the outer breadth of the wall (b) = inner breadth + 2d = 15 m + 2 × 0.1 m = 15.2 m Also, the height of the wall (h) = 4 m Now,cross-sectional area of the wall (A) = 2d(l + b – 2d) = 2 × 0.1(20 + 15.2 – 2 × 0.1) = 7 m2 ∴ Volume of the wall = cross-sectional Area × height = 7 m2 × 4 m = 28 m3 = 28 × 100 × 100 × 100 cm3 Now, volume of each brick = 20 cm × 10 cm × 5 cm ∴ Number of bricks required = volume of wall volume of each brick = 28 × 100 × 100 × 100 cm3 20 × 10 × 4 cm3 = 28,000 bricks Again, the cost of 1,000 bricks = Rs 18,000 the cost of 1 brick = Rs 18,000 1,000 the cost of 28,000 bricks = Rs 18,000 1,000 × 28,000 = Rs 5,04,000 Also, labourer wages for 2 labourers per day = 2 × Rs 650 = Rs 1,300 Labourer wages for 2 labourers for 5 days = 5 × Rs 1,300 = Rs 6,500 ∴ The total estimation of building the compound wall = Rs 5,04,000 + Rs 6,500 = Rs 5,10,500 4 m 10 cm 15 m 20 m

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 127 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids Example 4: A rural municipality built a water tank in a ward for supplying the water to the villagers as shown in the figure. The walls of the tank are rectangular. The inner diagonal AC of its floor ABCD is 12 m and perpendicular lengths DM and BN from the opposite corners D and B to the diagonal AC are 6 m and 8 m respectively and the tank is 5 m high. (i) What is the area of the floor of the tank? (ii) What is the capacity of the tank? (iii) How much amount will be collected from the full tank of water at Rs 50 per 10000 litres? (iv) If the villagers consume 60,000 litres of water per day, for how many days would the water of the full tank be sufficient? Solution: Here, AC = 12 m, DM ⊥ AC and BN ⊥ AC, DM = 6 m, BN = 8 m and AE (h) = 5 m. (i) Area of the floor of the tank = Area of ∆ACD + Area of ∆ABC = 1 2 AC × DM + 12 AC × BN = 1 2 AC (DM + BN) = 1 2 × 12 (6 + 8) m2 = 84 m2 (ii) The volume of the tank = Area of base × height = 84 × 5 m3 = 420 m3 We know, 1 m3 = 1000 litres ∴ 420 m3 = 420 × 1000 litres = 4,20,000 litres Thus, the capacity of the tank is 4,20,000 litres. (iii) The cost of 10000 litres of water = Rs 50 or, The cost of 1 litres of water = Rs 50 10000 or, The cost of 4,20,000 litres of water = Rs 50 10000 × 4,20,000 = Rs 2,100 Thus, Rs 2,100 would be collected from the full tank of water. (iv) 60,000 litres of water can be consumed in 1 day. or, 1 litre of water can be consumed in 1 60000 days. or, 4,20,000 litre of water can be consumed in 1 60000 × 4,20,000 = 7 days. Thus, a full tank of water can be lasted in 7 days. H G F C E D A M B N

Vedanta Excel in Mathematics - Book 10 128 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids EXERCISE 6.3 General section 1. a) If A is the area of the floor of a room and R be the rate of cost of carpeting the floor, write the formula to find the total cost (T) of carpeting the floor. b) If A be the total area of a 4 walls and ceiling of a room and T be the total cost of colouring the walls and ceiling of the room, write the formula to find the rate of the cost (R) of colouring. c) If R be the rate of cost and T be the total cost of plastering the floor of a room, write the formula to find the area (A) of the floor. 2. a) If V be the volume of a wall and v be the volume of a brick, write the formula to find the number of bricks (N) required to build the wall. b) If V be the volume of a wall and N be the number of bricks required to build the wall, write the formula to find the volume of each brick. c) If v be the volume of a brick and N be the number of bricks required to build a wall, write the formula to find the volume of the wall. 3. a) The area of the floor of a room is 35 m2 . Find the cost of carpeting the floor at the rate of Rs 90 per sq. m. b) A room is 8 m long and 5 m broad. Find the cost of carpeting its floor at Rs 75 per sq. m. c) A rectangular room is 9 m long, 6 m broad and 5 m high. Find the cost of colouring its 4 walls and ceiling at the rate of Rs 150 per sq. m. 4. a) The area of square base of a water tank is 4 m2 and the height of the tank is 2.5 m. Find the capacity of the tank in litre. b) The internal length of a square base water tank is 2 m and its height is 1.5 m. How many litres of water does it hold when its is full? c) The area of the circular base of a cylindrical water tank is 6.16 m2 and it is 1.5 m high. What is the capacity of the tank in litre? d) The internal diameter of a cylindrical water tank is 2 m and it is 3.5 m high. How many litres of water does it hold when it is full? Creative section 5. a) The adjoining figure represents the kitchen garden of a house. Estimate the cost of fencing the garden with 7 rounds of fencing wire. Rate of cost per metre of wire is Rs 30. Labourer wages is Rs 650 per labourer per day and the completion of the fencing wire requires 3 days for 2 labourers. b) The figure alongside is a compound of a house. The length of the diagonal AC = 40 m, DP = 25 m and BQ = 10 m. Estimate the cost of plastering the compound at the rate of Rs 75 per sq. m. A B C D 16 m 14 m 20 m 10 m A B C D P Q

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 129 Vedanta Excel in Mathematics - Book 10 Mensuration (II): Combined Solids 6. a) The dimension of a rectangular room is 10 m long, 8 m broad, and 5 m high. (i) Estimate the cost of carpeting its floor at Rs 90 per sq.m with Rs 600 labourer charge. (ii) Estimate the cost of colouring its walls and ceiling at Rs 105 per sq.m. b) A rectangular room is 8 m long, 7 m broad and 4.5 m high. Estimate the cost of plastering its floor, walls and ceiling at the rate of Rs 110 per sq.m. 7. The adjoining figure is a rectangular compound wall of a house. Estimate the cost of building the walls by using bricks of size 20 cm × 8 cm × 5 cm. The rate of the cost of bricks is Rs 1,750 per 1,000 bricks. Labourer wages is Rs 750 per labourer per day and 3 labourers are required for 4 days. 8. In a rural municipality, a water tank is constructed in a ward for supplying the water to the villagers as shown in the figure. The walls of the tank are rectangular. The inner diagonal AC of its floor ABCD is 10 m and perpendicular lengths DM and BN from the opposite corners D and B to the diagonal AC are 6 m and 7 m respectively and the tank is 5 m high. (i) What is the area of the floor of the tank? (ii) What is the capacity of the tank? (iii) How much amount will be collected from the full tank of water at Rs. 50 per 10000 litres? (iv) If the villagers consume 25,000 litres of water per day, in how many days would the water of the full tank be sufficient? Project Work and Activity Section 9. a) Measure the length and breadth of your school compound. Calculate its perimeter and area. Then, (i) Estimate the cost of fencing the compound according to the present cost of materials and labourer wages. (ii) Estimate the cost of plastering the compound according to the present cost of materials and labourer wages. b) Measure the length, breadth and height of your classroom. Calculate the area of (i) floor, (ii) 4 walls and (iii) ceiling. Estimate the cost of plastering the floor, walls and ceiling at the usual rate of cost in your locality. Also, estimate the cost of colouring the walls and ceiling. 5 m 12 cm 25 m 32 m H G F C E D A M B N

Vedanta Excel in Mathematics - Book 10 130 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (II): Combined Solids Assessment –III 1. Nepal Model Secondary School built the Saraswoti temple in its premises. The temple is the combination of the shapes of prism and pyramid with equal base. The dimensions of the prism are 8 m × 8 m × 4 m with four doors each of size 3 m × 1.5 m on the sides of the walls. The total height of the temple is 7 m. Answer the following questions on the basis of the information given above. (a) Write down the formula for calculating the volume of square based pyramid with side length ‘a’ units and slant height ‘l ‘units. (b) Find the slant height of the roof of the temple. (c) Find the cost for the zinc sheet to cover its roof at Rs 125 per square meter. (d) Compare the cost of covering the roof with zinc sheet and the colouring its outer walls at Rs 120 per square meter. 2. There are hollow transparent math kits in the shape of cone and square based pyramid made with plastic in mathematics laboratory. The slant height of both the kits is equal. The conical kit with slant height 25 cm has a total surface area of 704 cm2 . (a) The formula to find the volume of a cone having base radius ‘r’ units, height ‘h’ units and slant height ‘l’ units is: (i) πr2 h (ii) πr (r + l) (iii) πrl (iv) 13 πr2 h (b) Find the radius of the cone. (c) If the length of each side of base of pyramid kit is equal to the diameter of the conical kit, which kit can hold more water and by much? 3. A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Answer the following questions. (a) The radius of hemispherical part is equal to: (i) diameter of cylindrical part (ii) diameter of hemispherical part. (iii) height of cylindrical part (iv) height of the hemispherical part. (b) Find the capacity of the vessel. (c) Find the inner surface area of the vessel. 4. Due to floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of the same diameter and height 2.8 m. The canvas to be used to make the tent costs Rs 150 per square metre. Answer the following questions. (a) How many square metre of canvas is required for the tents? (b) Find the amount the associations will have to pay. 13 cm 14 cm

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 131 Vedanta Excel in Mathematics - Book 10 Unit 7 Sequence and Series Classwork - Exercise 7.1 Sequence and Series – Looking back 1. Let’s think and insert next two terms of each of the following sequences. (a) The next two terms of the sequence 2, 5, 8, 11, 14, are …...... and …...... (b) The next two terms of the sequence 3, 6, 9, 12, 15 are …...... and …...... (c) The next two terms of the sequence 41, 36, 31, 26 are …...... and …...... (d) The next two terms of the sequence 2, 4, 8, 16, 32 are …...... and …...... (e) The next two terms of the sequence 320, 160, 80, 40 are …...... and …...... 2. Let’s tick (√) the correct option from the bracket. (a) 5, 11, 17, 23, 29, 35, ….., 53 is a/an … sequence. (finite, infinite) (b) 4, 12, 36, 108, 324,…….. is a/an … sequence. (finite, infinite) (c) 7, 14, 28, 56, 112, 224 is a/an … progression. (arithmetic, geometric) (d) 96, 92, 88, 84, 80, 76 is a/an … progression. (arithmetic, geometric) 7.2 Arithmetic sequence (A.S.) Suppose, the distance of each rung of the ladder from its foot are 25 cm, 50 cm, 75 cm, ….. (i) What is the distance between any two consecutive rungs in the ladder? (ii) What is the distance of the fourth rung from the foot? Here, the distance between any of two consecutive rungs is constant. Thus, the distances of the rungs of ladder measured from its foot form an arithmetic sequence. A sequence is said to be arithmetic if its terms are increased or decreased by the same number. In arithmetic sequence every term after the first term is the sum of preceding term and a fixed number, called the common difference of the sequence. For example: (i) 5, 9, 13, 17, ….., (ii) 33, 30, 27, 24, …. are arithmetic sequence (A.S.) or arithmetic progression (A.P.)

Vedanta Excel in Mathematics - Book 10 132 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Facts to remember 1. An ordered list of numbers following a definite rule is known as a sequence. 2. A sequence in which the difference between any two consecutive terms remains constant is known as arithmetic sequence (A.S.) or arithmetic progression (A.P.) 3. Common difference (d) = t 2 – t1 or t 3 – t2 or t 4 – t3 4. In an A.P. with the first term a and common difference d, the nth term (t n) = a + (n – 1) d 5. If pth and qth terms of an A.P. are x and y respectively, then, common difference, d = x – y p – q 7.3 Arithmetic mean between two numbers Let’s consider an arithmetic sequence 10, 15, 20. Here, 15 is called arithmetic mean between 10 and 20. Thus, for an A.P. having three terms, the middle term is called the arithmetic mean (A.M.) between the first and the third terms. Let three numbers a, m and b are in an arithmetic sequence. Then, by definition, we have m – a = b – m or, 2m = a + b or, m = a + b 2 Hence, the arithmetic mean between two numbers a and b is given by A.M. = a + b 2 7.4 Arithmetic means between two numbers Let’s take an arithmetic sequence 3, 7, 11, 15, 19. Here, the terms 7, 11 and 15 are called arithmetic means between 3 and 19. The terms between the first term and the last term of an A.P. are called the arithmetic means (A.M.’s). Let a and b be the first and the last terms of an A.P., and m1 , m2 , ...... mn are n arithmetic means between a and b. Then a, m1 , m2 ...... mn, b, are in A.P. The number of means = n The total number of terms in the A.P. (N) = n + 2 Then, last term (t n) = a + (N – 1) d or, b = a + [(n+2) – 1] d or, b – a = (n + 2 – 1) d or, b – a = (n + 1)d ∴ d = b – a n + 1 Again, m1 = t2 = a + d, m2 = t3 = a + 2d, m3 = t4 = a + 3d, …, mn = a + nd www.geogebra.org/classroom/kyvsw2x6 Classroom code: KYVS W2X6 Vedanta ICT Corner Please! Scan this QR code or browse the link given below: First mean 3 t 1 7 t 2 11 t 3 15 t 4 19 t 5 Second mean Third mean

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 133 Vedanta Excel in Mathematics - Book 10 Sequence and Series Facts to remember 1. A.M. between any two numbers a and b = a + b 2 2. If there are n arithmetic means between any two numbers a and b, then, common difference is obtained by the formula, d = b – a n + 1 3. 1st mean (m1 ) = 2nd term = a + d, 2nd mean (m2 ) = 3rd term = a + 2d, ……. 4. Last mean or nth mean (mn) = a + nd or, b – d Worked-out Examples Example 1: Find the arithmetic mean between 11 and 33. Solution: Here, first term (a) = 11 and last term (b) = 33 Now, A.M. = a + b 2 = 11 + 33 2 = 44 2 = 22 Hence, the arithmetic mean between 11 and 33 is 22. Example 2: The weights of a girl and a boy are in the ratio of 3:4. If their arithmetic mean weight is 56 kg, find their weights. Solution: Let, the weight of girl, a = 3x and the weight of boy, b = 4x Now, A.M. = a + b 2 or, 56 = 3x + 4x 2 or, 7x = 112 or, x = 112 7 ∴ x = 16, Then, 3x = 3 × 16 = 48 and 4x = 4 × 16 = 64 Hence, the weight of the girl is 48 kg and the weight of the boy is 64 kg. Example 3: Insert 5 arithmetic means between 2 and 50. Solution: Here, first term (a) = 2, last term (b) = 20 and number of means (n) = 5 Now, common difference (d) = b – a n + 1 = 20 – 2 5 + 1 = 18 6 = 3 Again, m1 = a + d = 2 + 3 = 5 m2 = a + 2d = 2 + 2 × 3 = 8 m3 = a + 3d = 2 + 3 × 3 = 11 m4 = a + 4d = 2 + 4 × 3 = 14 m5 = a + 5d = 2 + 5 × 3 = 17 Hence, 5 arithmetic means between 2 and 20 are 5, 8, 11, 14 and 17. Checking: Ratio = 48: 64 = 48 64 = 3:4 A.M. = a + b 2 = 48 + 64 2 = 56, which are given in the question. Alternative process of finding d: No. of terms (n) = 5 + 2 = 7 Now, seventh term (t7 ) = 20 or, a + 6d = 20 or, 2 + 6d = 20 or, 6d = 18 ∴ d = 3

Vedanta Excel in Mathematics - Book 10 134 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Example 4 If 8, x, y, z, – 8 are in A.P., find: (i) the common difference (ii) the values of x, y and z. Solution: Here, first term (a) = 8, last term (b) = – 8 Number of means (n) = 3 Now, common difference (d) = b – a n + 1 = –8 – 8 3 + 1 = –16 4 = –4 (i) Again, x = m1 = a + d = 8 + (– 4) = 8 – 4 = 4 y = m2 = a + 2d = 8 + 2 (– 4) = 8 – 8 = 0 z = m3 = a + 3d = 8 + 3 (– 4) = 8 – 12 = – 4 Hence, x = 4, y = 0 and z = – 4. Example 5: There are n arithmetic means between 2 and 22. If the third mean is 14, find: (i) common difference (ii) the value of n (iii) the remaining means Solution: Let m1 , m2 , m3 , …, mn be the n arithmetic means between 2 and 22 respectively. Then, 2, m1 , m2 , m3 , …, 22 is an A.P. Here, first term (a) = 2, last term (b) = 22 and the third mean (m3 ) = 14 (i) Now,the third mean (m3 ) = a + 3d or, 14 = 2 + 3d or, 12 = 3d or, 4 = d Hence, the common difference is 4. Also, d = b – a n + 1 or, 4 = 22 – 2 n + 1 or, 4n + 4 = 20 or, n = 4 Hence, there are 4 arithmetic means. (ii) Again, m1 = a + d = 2 + 4 = 6 m2 = a + 2d = 2 + 2 × 4 = 10 m4 = a + 4d = 2 + 4 × 4 = 18 Hence, the remaining means are 6, 10 and 18. Example 6: There are n arithmetic means between 10 and 50. If the ratio of first mean to last mean is 1: 3, find: (i) common difference (ii) the value of n (iii) the means Solution: Let m1 , m2 , m3 , …, mn be the n arithmetic means between 10 and 50 respectively. Then, 10, m1 , m2 , m3 , …, 50 is an A.P. Here, first term (a) = 10, last term (b) = 50 Alternative process for finding d: No. of terms (n) = 3 + 2 = 5 Now, seventh term (t5 ) = – 8 or, a + 4d = – 8 or, 8 + 4d = – 8 or, 4d = – 16 ∴ d = – 4

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 135 Vedanta Excel in Mathematics - Book 10 Sequence and Series (i) Now, first mean (m1 ) = a + d = 10 + d and last mean (mn) = b – d = 50 – d By question, m1 mn = 1 3 or, 10 + d 50 – d = 1 3 or, 30 + 3d = 50 – d or, 4d = 20 ∴ d = 5 (iii) Again, m1 = a + d = 10 + 5 = 15 m2 = a + 2d = 10 + 2 × 5 = 20 m3 = a + 3d = 10 + 3 × 5 = 25 m4 = a + 4d = 10 + 4 × 5 = 30 m5 = a + 5d = 10 + 5 × 5 = 35 m6 = a + 6d = 10 + 5 × 5 = 40 m7 = a + 7d = 10 + 5 × 5 = 45 Hence, 7 arithmetic means between 10 and 50 are 15, 20, 25, 30, 35, 40 and 45. Example 7: There are 5 arithmetic means between a and b. If the first mean is 8 and the third mean is 18, find: (i) the first term and common difference (ii) the last term (iii) the remaining means Solution: Here, first term = a, last term = b and number of arithmetic means (n) = 5 (i) Now, first mean (m1 ) = 8 or, a + d = 8 ∴a = 8 – d … (i) And, third mean (m3 ) = 18 ∴a + 3d = 18 … (ii) Putting the value of a from equation (i) in equation (ii), we get, (8 – d) + 3d = 18 or, 2d = 10 ∴ d = 5 Putting the value of d in equation (i), we get ∴a = 8 – 5 = 3 Hence, the first term is 3 and common difference is 5. (ii) Also, d = b – a n + 1 or, 5 = b – a 5 + 1 or, b – 3 = 30 ∴ b = 33 Hence, the last term is 33. (iii) Again, remaining means are: m2 = a + 2d = 3 + 2 × 5 = 13 m4 = a + 4d = 3 + 4 × 5 = 23 m5 = a + 5d = 3 + 5 × 5 = 28 Hence, the remaining means are 13, 23 and 28. (ii) Also, d = b – a n + 1 or, 5 = 50 – 10 n + 1 or, 5n + 5 = 40 or, 5n = 35 ∴ n = 7

Vedanta Excel in Mathematics - Book 10 136 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Example 8: A group of Canadian tourists came to Nepal to climb up the Mt. Annapurna. They climbed up the mountain only 6 hours a day. On a day, they climbed up 600 m in the first hour and a certain distance less than the previous hour in each subsequent hour. In this way, they climbed up 500 m in the 6th hour. (i) By how many meters did they lower the distance of climbing up the mountain in each hour? (ii) How many distances did they climb in each hour between the first and last hours? Solution: Let m1 , m2 , m3 and m4 be the distances of climbing the mountain in the 2nd, 3rd, 4th and 5th hours respectively. Then, 600, m1 , m2 , m3 , m4 , 500 are in an A.P. Here, first term (a) = 600 m, last term (b) = 500 m Number of means (n) = 4 (i) Now, common difference (d) = b – a n + 1 = 500 – 600 4 + 1 = –100 5 = –20 So, they climbed 20 m less than the previous hour in each subsequent hour. (ii) Again, m1 = a + d = 600 + (– 20) = 580 m2 = a + 2d = 600 + 2 (– 20) = 560 m3 = a + 3d = 600 + 3 (– 20) = 540 m4 = a + 4d = 600 + 4 (– 20) = 520 Hence, they climbed 580 m, 560 m, 540 m and 520 m in each hour between the first and last hour of the day. EXERCISE 7.1 General section 1. a) Define arithmetic mean. b) What is the arithmetic mean between any two numbers p and q? c) What is the arithmetic mean between 5 and 7? 2. a) If there are ‘n’ arithmetic means between the numbers a and b, what will be the common difference (d)? b) If there are 9 arithmetic means between 5 and 55, what is the common difference (d)? 3. a) In an A.S. if there are 5 means between a and b, how many terms are there? b) What is the second mean in the arithmetic sequence 4, 7, 10, 13, 16, 19, 22? c) What is the last mean in the arithmetic sequence –7, – 2, 3, 8, 13, 18? d) An A.P. has first term a, common difference d and n number of means, what is the last mean? Alternatively, No. of terms (n) = 6 Now, sixth term (t6 ) = 500 or, a + 5d = 500 or, 600 + 5d = 500 or, 5d = – 100 ∴ d = – 20 Then, find t2 , t3 , …, t5

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 137 Vedanta Excel in Mathematics - Book 10 Sequence and Series 4. Find the arithmetic mean between the following two numbers. a) 7 and 13 b) –9 and 45 c) 30 and – 30 d) – 53 and – 35 e) 7 2 and 9 2 f) 5 2 and 5 3 g) (a2 + b2 ) and (a2 – b2 ) h) (a – b) 2 and (a + b) 2 5. Find the 2nd term from the following arithmetic sequences. a) 1st term = 25 and 3rd term = 75 b) 1st term = – 5 and 3rd term = – 55 6. Find the term as indicated in the following arithmetic sequences. a) 4th term = 11, 6th term = 17, find the 5th term. b) 8th term =– 6, 10th term = 12, find the 9th term. c) 5th term = 9, 7th term = 19, find the 6th term. d) 9th term =– 4, 11th term = – 44, find the 10th term. 7. Find the value of x in the following conditions. a) 7, x, 11 are in an A.P. b) 10, x, 20 are in an A.P. c) 13, x, – 31 are in an A.P. 8. a) If the arithmetic mean between the numbers x and 8 is 5, find the value of x. b) If 15 is the arithmetic mean between the numbers 5 and y, find the value of y. c) If the arithmetic mean between 3x and 4x is 14, find the numbers. d) One number is three times the other and their arithmetic mean is 20, find the numbers. e) The heights of a tree and a temple are in the ratio 2:3 and their arithmetic mean height is 15 m, find their heights. Creative section–A 9. Insert the arithmetic means between the given terms. a) 3 arithmetic means between 6 and 22. b) 4 arithmetic means between 10 and 25. c) 5 arithmetic means between – 3 and 33. d) 6 arithmetic means between – 7 and – 77. 10. a) If 13, x, y, 28 are in A.P., find: (i) the common difference (ii) the values of x and y. b) If 96, m, n, 69 are in A.P., find: (i) the common difference (ii) the values of m and n. c) If 7, p, q, r, 35 are in A.P., find: (i) the common difference (ii) the values of p, q and r. d) If – 2, a, b, c, 66 are in A.P., find: (i) the common difference (ii) the values of a, b and c. 11. a) There are n arithmetic means between 7 and 27. If the second mean is 15, find: (i) the common difference (ii) the value of n (iii) the remaining means b) There are n arithmetic means between 24 and 64. If the fourth mean is 56, find: (i) the common difference (ii) the value of n (iii) the remaining means 12. a) There are n arithmetic means between 1 and 19. If the ratio of the first mean to last mean is 1: 4, find: (i) the common difference (ii) the value of n (iii) the means

Vedanta Excel in Mathematics - Book 10 138 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series b) n arithmetic means are inserted between 20 and 80. If the ratio of the first mean to last mean is 1: 3, find: (i) the common difference (ii) the value of n (iii) the means c) In an A.P., a few arithmetic means are inserted between 75 and 105. If the ratio of first mean to last mean is 4: 5, find: (i) the common difference (ii) the numbers of means (iii) the means 13. a) There are 5 arithmetic means between a and b. If the first and third means are 5 and 11 respectively, find: (i) the first term a and the last term b (ii) the remaining means b) There are 6 arithmetic means between a and b. If the first and fourth means are 3 and 15 respectively, find: (i) the first term a and the last term b (ii) the remaining means Creative section–B 14. a) Saroj has bought a new Korean language book. He decided to learn new Korean words every day so that the number of new words that he learned each day forms an arithmetic sequence. In the first day, he learned 10 new words and in the seventh day, he learned 100 new words. (i) By how many more new words did he learn in each day than the preceding day? (ii) How many new words did he learn in each day from second to sixth days? b) The population of a municipality increased by 500 in the year B.S. 2079. The rate of population growth is expected to decrease by a number every year so that there will be the increase of only 400 people in B.S. 2084. (i) What is the expected decrease rate in population growth per year? (ii) What will be the increased number of population in each year from B.S. 2080 to B.S. 2083? 7.5 Sum of first n terms of an Arithmetic Series The German mathematician Carl Friedrich Gauss (1777−1855) was a child prodigy, able to correct his father’s arithmetic at the age of three. One day, his teacher asked to add up all the numbers from 1 to 100. Gauss was only seven years old, but within minutes he had the answer: the first one hundred natural numbers add up to 5050. He had used the following pattern to find the sum. S = 1 + 2 + 3 + 4 + … + 98 + 99 + 100 S = 100 + 99 + 98 + 97 + … + 3 + 2 + 1 This gives 100 pairs of numbers that add up to 101. But the series has been used twice, ∴ S100 = 1 2 × 100 × 101 = 5050. Gauss became one of the most influential and innovative mathematicians of his time and his ideas still inspire mathematicians today.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 139 Vedanta Excel in Mathematics - Book 10 Sequence and Series For an A.P., let a = first term, d = common difference, l = last term and n = number of terms. Then, first n terms are: a, a + d, a + 2d, a + 3d, ....., l – 3d, l – 2d, l – d, l Let Sn denote the sum of n terms of the A.P. Now, Sn = a + (a + d) + (a + 2d) + ...... + (l – 2d) + (l – d) + l … (i) On writing the terms in reverse order, Sn = l + (l – d) + (l – 2d) + ..... + (a + 2d) + (a + d) + a … (ii) Adding the corresponding terms of (i) and (ii), we get Sn = a + (a + d) + (a + 2d) + ...... + (l – 2d) + (l – d) + l Sn = l + (l – d) + (l – 2d) + ..... + (a + 2d) + (a + d) + a 2Sn = (a + l) + (a + l) + (a + l) + ....... + (a + l) = Sum of n times (a + l) = n (a + l) ∴Sn = n(a + l) 2 We know, the last term (l) = a + (n – 1) d ∴Sn = n(a + l) 2 = n[a + {a + (n – 1)d}] 2 = n 2 [2a + (n – 1) d] Facts to remember When a, l and n are known, Sn = n(a + l) 2 When a, d and n are known, Sn = n 2 [2a + (n – 1) d] Corollary 1: The sum of first n natural numbers is given by Sn = n 2 (n + 1) Proof: Since, the numbers 1, 2, 3, …., n are the first n natural numbers. They form an arithmetic sequence with first term (a) = 1, common difference (d) = 2 – 1 = 1 and number of terms (n) = n Now, Sn = n 2 [2a + (n – 1) d] = n 2 [2 × 1 + (n – 1) × 1] = n 2 (2 + (n – 1 ) = n 2 (n + 1) Hence, the sum of first n natural number is given by Sn = n 2 (n + 1) . Corollary 2: The sum of first n odd numbers is given by Sn = n2 Proof: Since, the numbers 1, 3, 5, …., 2n – 1 are the first n natural odd numbers. They form an arithmetic sequence with first term (a) = 1, common difference (d) = 2 Now, Sn = n 2 [2a + (n – 1) d] = n 2 [2 × 1 + (n – 1) × 2] = n 2 (2 + 2n – 2) = n 2 × 2n = n2 Hence, the sum of first n odd number is given by Sn = n2 . www.geogebra.org/classroom/anvngbkm Classroom code: ANVN GBKM Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Alternatively, First term (a) = 1, last term (l) = n No. of terms (n) = n ∴ Sn = n 2 (a + l) = n 2 (1+ n) = n 2 (n + 1) Alternatively, First term (a) = 1, last term (l) = 2n – 1 No. of terms (n) = n ∴ Sn= n 2 (a + l) = n 2 (1+ 2n – 1) = n2

Vedanta Excel in Mathematics - Book 10 140 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Corollary 3: The sum of first n even numbers is given by Sn = n (n + 1) Proof: Since, the numbers 2, 4, 6, …., 2n are the first n natural even numbers. They form an arithmetic sequence with first term (a) = 2, common difference (d) = 2 and number of terms (n) = n Now, Sn = n 2 [2a + (n – 1) d] = n 2 [2 × 2 + (n – 1) × 2] = n 2 (4 + 2n – 2) = n 2 (2n + 2) = n (n + 1) Hence, the sum of first n odd natural number is given by Sn = n (n + 1) Worked-out Examples Example 1: Find the sum of first 10 terms of an arithmetic series: – 8 – 4 + 0 + 4 + 8, … Solution: Here, first term (a) = – 8, common difference (d) =– 4 – (–8) = – 4 + 8 = 4 and no. of terms (n) = 10 Sum of first 10 terms (S10) =? Now, Sn = n 2 [2a + (n – 1) d] = 10 2 [2 (– 8)+ (10 – 1) 4] = 5 (– 16 + 9 × 4) = 5 × 20 = 100 Hence, the required sum is 100. Example 2: Find the sum of first 20 odd numbers. Solution: Here, the given series is 1 + 3 + 5 + …; first term (a) = 1, common difference (d) = 2, number of terms (n) = 20, sum of first 20 terms (S20) =? Now, Sn = n 2 [2a + (n – 1) d] = 20 2 [2 × 1 + (20 – 1) 2] = 10 (2 + 19 × 2) = 10 × 40 = 400 Hence, the required sum is 400. Example 3: For the A.P., 3 + 11 + 19 + … + 99, find: (i) number of terms (ii) the sum of all the terms Solution: Here, first term (a) = 3, common difference (d) = 11 – 3= 8, last term (l) = 99 (i) Now, last term (tn) = a + (n – 1)d or, 99 = 3 + (n – 1) 8 Alternatively, First term (a) = 2, last term (l) = 2n No. of terms (n) = n ∴ Sn = n 2 (a + l) = n 2 (2+ 2n) = n 2 × 2(n + 1) = n (n + 1) Alternatively, Last term (l) = t10 = a + 9d = –8 + 9 × 4 = 28 ∴ Sn = n 2 (a + l) = 10 2 (–8 + 28) = 5 × 20 = 100 Alternatively, Sum of first 20 odd numbers (Sn) = n2 = 202 = 400

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 141 Vedanta Excel in Mathematics - Book 10 Sequence and Series or, 99 = 3 + 8n – 8 or, 104 = 8n or, n = 13 Thus, the number of terms is 13. (ii) Again, Sn = n 2 (a + l) = 13 2 (3 + 99) = 13 2 × 102 = 663 Hence, the required sum is 663. Example 4: In an arithmetic series, the first term is 5, the last term is 50 and the sum of all the terms is 1100, find the number of terms. Solution: Here, first term (a) = 5, last term (l) = 50 and sum of terms (Sn) = 1100 Now, Sn = n 2 (a + l ) or, 1100 = n 2 (5 + 50) or, 2200 = 55n or, n = 40 Hence, there are 40 terms in the series. Example 5: If the 10th term of an A.P. is 24, find the sum of the first 19 terms. Solution: Let, a and d be the first term and common difference of an A.P. Then, 10th term (t10) = 24 ∴a + 9d = 24 … (i) Now, Sn = n 2 [2a + (n – 1) d] ∴ S19 = 19 2 [2a + (19 – 1) d] = 19 2 (2a + 18 d) = 19 2 × 2(a + 9d) = 19 × 24 [From (i), a + 9d = 24] = 456 Hence, the required sum is 456. Example 6: Given arithmetic series is 52 + 48 + 44 + 40 + … (i) Find the sum of the first 10 terms of the series. (ii) If the sum of the first n terms of the series is 360, find the possible values of n. (iii) Explain the existence of two values of n. Solution: Here, first term (a) = 52, common difference (d) =48 – 52 = – 4 (i) Sum of the first 10 term (S10) = n 2 [2a + (n – 1) d] = 10 2 [2 × 52 + (10 – 1) (–4)] = 5 (104 – 36) = 5 × 68 = 340 (ii) Sum of the first n terms (Sn) =360 or, n 2 [2a + (n – 1) d] =360 Checking: When n = 13, last term (t13) = a + 12d = 3 + 12 × 8 = 99, which is given in the question. Checking: S40 = 40 2 (5 + 50) =20 × 55 = 1100, which is given in the question. Checking: S19 = 19 2 [2a + (19 – 1) d] or, 456 =19 2 × 2(a + 9d) or, a + 9d = 24 i.e., 10th term = 24, which is given in the question.

Vedanta Excel in Mathematics - Book 10 142 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series or, n 2 [2 × 52 + (n – 1) (–4)] = 360 or, n [104 – (n – 1) 4] = 720 or, n [104 – 4n + 4] = 720 or, 108n – 4n2 = 720 or, 4 (27n – n2 ) = 720 or, n2 – 27n + 180 = 0 or, n2 – (15 + 12) n + 180 = 0 or, n2 – 15n – 12n + 180 = 0 or, n(n– 15) – 12(n – 15) = 0 or, (n– 15) (n – 12) = 0 Either, n – 15 = 0 ∴n = 15 Or, n – 12 = 0 ∴n = 12 (iii) t 13 = a + 12d = 52 + 12 (– 4) = 52 – 48 = 4 t 14 = a + 13d = 52 + 13 (– 4) = 52 – 52 = 0 t 15 = a + 14d = 52 + 14 (– 4) = 52 – 56 = – 4 Since, the sum of 13th, 14th and 15th terms is zero. So, the sum of the first 12 terms is same as the sum of the first 15 terms. Hence, the existence of two values of n shows that there are two A.P.s satisfying the condition. Example 7: The fourth and the sixth terms of an arithmetic series are 17 and 25 respectively. Find: (i) the first term and common difference (ii) the sum of first 10 terms. Solution: (i) Let a and d be the first term and the common difference of an AP respectively. Then, 4th term (t4 ) = 17 ∴a + 3d = 17 … (i) Also, 6th term (t6 ) = 25 ∴a +5d = 25 … (ii) Now, subtracting (i) from (ii), we get a + 5d = 25 a + 3d = 17 (–) (–) (–) 2d = 8 ∴ d = 4 Also, putting the value of d in (i), we get a + 3×4 = 17 ∴ a = 17 – 12 = 5 Hence, the required first term is 5 and the common difference is 4. (ii) Again, we know, Sn = n 2 [2a + (n – 1) d] ∴Sum of first 10 terms (S10) = 10 2 [2 × 5 + (10 – 1) 4] = 5 (10 + 9 × 4) = 5 × 46 = 230 Hence, the sum of the first 10 terms is 230. Checking: When n = 15, Sn = n 2 [2a + (n – 1) d] = 15 2 [2 × 52 + (15– 1) (–4)] = 15 2 ×48 = 360, which is given. When n = 12, Sn = n 2 [2a + (n – 1) d] = 12 2 [2 × 52 + (12– 1) (–4)] = 6 × 60 = 360, which is given. Checking: Fourth term (t4 ) = a + 3d = 5 + 3 × 4 = 17 Sixth term (t6 ) = a + 5d = 5 + 5 × 4 = 25, which are given in the question.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 143 Vedanta Excel in Mathematics - Book 10 Sequence and Series Example 8: In an A.P., the 5th term is treble the second term and the sum of the first ten terms is 50 find: (i) the first term and common difference (ii) the sum of the first 20 terms. Solution: (i) Let a and d be the first term and the common difference of an AP respectively. Then, 5th term (t5 ) = 3 × 2nd term (t2 ) or, a + 4d = 3 (a + d) or, d = 2a … (i) Also, sum of first ten terms (S10) = 50, number terms (n) = 10 We know; Sn = n 2 [2a + (n – 1) d] or, 50 = 10 2 [2a + (10 – 1) d] or, 50 = 5 (2a + 9d) ∴ 2a +9d = 10 … (ii) Now, putting the value of d in (ii), we get 2a +9 × 2a = 10 or, 20a = 10 ∴ a = 1 2 Also, putting the value of a in (i), we get d = 2 × 1 2 = 1 (ii) Sum of the first 20 terms (S20) =? We know, Sn = n 2 [2a + (n – 1) d] ∴Sum of the first 20 terms (S20) = 20 2 [2 × 1 2 + (20 – 1) 1] = 10 (1 + 19) =200 Hence, the sum of the first 20 terms is 200. Example 9: Mrs. Subba has just joined a job. She decides to save Rs 2,000 in her first month of work, Rs 2,250 in the second month, Rs 2,500 in the third month, and so on, forming an A.P. How much amount will she save in a year? Solution: Here, saving of first month (a) = Rs. 2,000, Increment in monthly savings (d) = Rs. 2,250 – Rs. 2,000 = Rs. 250 Number of months (n) = 12, total saving in a year (Sn) =? Now, Sn = n 2 [2a + (n – 1) d] ∴ S12 = 12 2 [2 × 2000 + (12 – 1) × 250] = 6 (4,000 + 2750) = 6 × 6,750 = 40,500 Hence, she will save Rs 40,500 in a year. Example 11: In a school, the ages of the students in Mathematics club are in AP, whose common difference is 4 months. If the youngest student is 10 years and the sum of the ages of all the students is 156 years, (i) How many students are there in the club? (ii) What is the age of eldest student in the class? Checking: S10 = 10 2 [2 × 1 2 + (10 – 1) 1] = 5 (1 + 9) =50, which is given in the question.

Vedanta Excel in Mathematics - Book 10 144 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Solution: Here, the age of youngest student (a) = 10 years, Common difference (d) = 4 months = 4 12 = 1 3 years Sum of ages (Sn) = 156 years. (i) Now, Sn = n 2 [2a + (n – 1) d] or, 156 = n 2 [2 × 10 + (n – 1) × 1 3] or, 312 = n ( 60 + n – 1 3 ) or, 936 = n2 + 59n or, n2 + 59n – 936 = 0 or, n2 +(72 – 13) n – 936 = 0 or, n2 + 72n – 13n – 936 = 0 or, n(n + 72) – 13(n + 72) = 0 or, (n + 72) (n – 13) = 0 ∴ n = – 72 or 13 But, the number of students cannot be negative. Hence, there are 13 students in the club. (ii) Again, tn = a + (n – 1)d = 10 + (13 – 1) × 1 3 = 10 + 4 = 14 Hence, the age of the eldest student in the club is 14 years. EXERCISE 7.2 General section 1. a) What is the sum of first n terms of an arithmetic series having first term a and last term l? b) What is the sum of first n terms of an arithmetic series having first term a and common difference d? c) What is the sum of first n natural numbers? d) What is the sum of first n odd numbers? e) What is the sum of first n even numbers? 2. Find the sum of following arithmetic series. a) 3 + 7 + 11+ … to 10 terms b) 4 + 11 + 18 + … to 15 terms c) – 6 + 3 + 12 + … to 12 terms d) – 2 – 6 –10 – 14 – … to 25 terms e) 12 + 32 + 52 +72+ … to 8 terms f) 1 2 + 1 + 3 2 + 2 + … to 16 terms 3. Find the sum of following series. a) 1 + 2 + 3 + … to 20 terms b) 1 + 2 + 3 + … to 100 terms c) 1 + 3 + 5 + … to 10 terms d) 1 + 3 + 5 + … to 30 terms e) 2 + 4 + 6 + … to 40 terms f) 2+ 4 + 6 + … to 55 terms 4. Find the sum of first 20 terms of each of the arithmetic sequence whose a) 1st term = 7 and 2nd term = 11 b) 1st term = 10 and 2nd term = 15 c) 1st term = 8 and 2nd term = 4 d) 1st term = – 6 and 2nd term = – 9 5. a) The first term of an arithmetic series having 20 terms is 4 and last term is 50, find the sum of the series. Checking: Sn = n 2 [2a + (n – 1) d] ∴ S13 = 13 2 [2 × 10 + (13 – 1) × 1 3] = 13 2 × 24 = 156, which is given in the question.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 145 Vedanta Excel in Mathematics - Book 10 Sequence and Series b) An AS has 30 terms and the 30th term is 65. If the first term is 5, find the sum of the series. c) There are 50 terms in an AP. If the first term is 20 and the 50th term is 120, find the sum of the series. d) The first term of an arithmetic progression is 10 and the common difference is 3, find the sum of its first 8 terms. e) The first term of an AP is –15 and the common difference is 9, find the sum of its first 25 terms. 6. a) In an arithmetic series; the first term is 4, the last term is 40 and the sum of all the terms is 1100, find the number of terms. b) The first term of an arithmetic progression is –7 and the last term is 85. If the sum of all the terms is 1248, how many terms are there? c) The sum of all the terms of an AP having 20 terms is 2200. If the first term is 20, what is its last term? d) The sum of all the terms of an AP having 30 terms is 750. If the last term is 105, what is its first term? 7. a) If the 3rd term of an arithmetic series is 13, find the sum of the first 5 terms. b) The 11th term of an AS is 10; find the sum of the first 21 terms. c) The sum of the first 11 terms of an AP is 1320; find the 6th term. d) If the sum of the first 55 terms of an AP adds up to 6600; find the 28th term. Creative section–A 8. a) The first term of an arithmetic series is 27 and the common difference is –3. (i) Find the sum of the first 20 terms of the series. (ii) If the sum of the first n terms of the series is 132, find the possible values of n. (iii) Explain the two values of n obtained in (ii). b) Given arithmetic series is 42 + 39 + 36 + 33 + … (i) Find the sum of the first 10 terms of the series. (ii) If the sum of the first n terms of the series is 315, find the possible values of n. (iii) Explain the double values of n. 9. a) The fifth term and the twelfth term of an arithmetic progression are 30 and 65 respectively. Find: (i) the first term and common difference (ii) the sum of the first 26 terms. b) The eighth term and the fifteenth term of an arithmetic series are 27 and 55 respectively. Find: (i) the first term and common difference (ii) the sum of the first 35 terms. 10. a) The sum of the first eight terms of an arithmetic series is 180 and its fifth term is five times of the first term, find: (i) the first term and common difference (ii) the sum of the first 10 terms. b) If the sum of the first ten terms of an arithmetic progression is 50 and its fifth term is treble of the second term, find: (i) the first term and common difference (ii) sum of the first twenty terms.

Vedanta Excel in Mathematics - Book 10 146 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series 11. a) The sum of the first seven terms of an arithmetic series is 14 and the sum of the first ten terms is 125. Find: (i) the first term and common difference (ii) the sum of the first 25 terms. b) If the sum of the first four terms of an arithmetic series is 26 and the sum of the first eight terms is 100, find (i) the first term and common difference (ii) the sum of the first 20 terms. c) The sum of the first 9 terms of an arithmetic series is 72 and the sum of the first 17 terms is 289, find the sum of the first 25 terms. Creative section–B 12. a) Mrs. Shakya accepts a position offered by a company with the negotiation of a salary of Rs 4,80,000 in the first year. She is guaranteed a raise of Rs 5,000 in each year. (i) Make the sequence of her yearly salaries. (ii) Calculate the total amount earned by her during 5 full years of employment. b) Mr. Sherpa is a farmer in Jumla. He sells the apples for a week in its season. If he sells 2 quintals of apples in the first day and sells 50 kg more apples in every subsequent day than the previous days; answer the following questions. (i) Make the sequence of quantities of apples in kg. (ii) How many quintals of apples does he sell during the week? 13. a) A wholesaler sold 3500 sacks of rice in the fourth year of his business and 5000 sacks of rice in the 7th year so that his annual sales are in A.P. (i) How many sacks of rice did he sell in the first year? (ii) By how many sacks of rice did he increase the sales in each year after the first year? (iii) How many sacks of rice did he sell in 10 years? b) A shoe factory produces 18000 pairs of shoes in its fifth year of production and 24000 pairs of shoes in eighth year. The increment in production remains constant throughout the production years. (i) How many pairs of shoes did it produce in the first year? (ii) By how many pairs of shoes is it increasing in its production in each year after the first year? (iii) How many pair of shoes will it produce in 15 years? 14. a) Minakshee is a 10 years old child. She makes a pattern with blocks on the floor. She puts two blocks in the first row, five in the second row, eight blocks in the third row and so on. (i) How many blocks are used up to the 7th row of the pattern? (ii) If only 100 blocks are available, what is the maximum number of rows of blocks that she can have in her pattern? b) Mr. Karki has just joined a job. He decides to save Rs 5,000 in his first month of work, Rs 5050 in the second month, Rs 5,100 in the third month, and so on, forming an A.P. (i) How much total amount will he save till 10th month? (ii) How long will he take to save a total of Rs 63,300?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 147 Vedanta Excel in Mathematics - Book 10 Sequence and Series Project Work and Activity Section 15. a) Make the groups of your friends. Write 1/1 question based on sum of the first any terms of the arithmetic series on the blank paper. Wad up the paper with question to make the snowballs by each groups and throw in front of the class. Tell your group leader to pick up any one of the snowballs. Solve the problem given in it in the group and present your solution in the class. b) Write the numbers between 102 and 201. Find the sum of the numbers which are not exactly divisible by 6. 7.6 Geometric sequence (G.S.) Every person has two parents (father and mother), 4 grand–parents and so on. Let’s observe the number of ancestors over the generations. Generation 1st 2nd 3rd 4th 5th … Number of ancestors 2 4 8 16 … … The sequence of ancestors in consecutive generations is 2, 4, 8, 16, …. Here, the number of ancestors of a generation is obtained by multiplying the number of ancestors of previous generation by a constant number. Thus, the number of ancestors over the generations forms a Geometric Sequence (G.S.). A sequence is said to be geometric if its terms are increased or decreased by the constant multiplication. In GS, every term after the first term is the product of preceding term and a fixed number, called the common ratio of the sequence. (i) 2, 6, 18, 54, ….., (ii) 400, 200, 100, 50, , …. are the examples of geometric sequences (G.S.) or geometric progressions (G.P.) Facts to remember 1. A sequence in which each term, except first term, is obtained on multiplying the preceding term by a fixed non–zero number is called a geometric sequence. The fixed number is called common ratio and denoted by r. 2. A sequence in which the ratios between any two consecutive terms are constant, is known as geometric sequence (G.S.) or geometric progression (G.P.) 3. Common ratio (r) = t 2 t 1 or t 3 t 2 or t 4 t 3 and so on. 4. The general term (nth) term, (tn) = arn–1, r ≠ 0 7.7 Geometric mean between two numbers Let’s consider a geometric sequence 3, 6, 12. Here, 6 is called geometric mean between 3 and 12. Thus, for a G.P. having three terms, the middle term is called the geometric mean (G.M.) between the first and the third terms. www.geogebra.org/classroom/kvunsu6t Classroom code: KVUN SU6T Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Let three numbers a, g and b are in a geometric sequence. Then, by definition, we have, g a = b g or, g2 = ab ∴ g = ab Hence, the geometric mean between two numbers a and b is given by G.M. = ab 7.8 Geometric means between two numbers Let’s take a geometric sequence 5, 10, 20, 40, 80. Here, the terms 10, 20 and 40 are called geometric means between 5 and 80. The terms between the first term and the last term of a G.P. are called the geometric means (G.M.’s). Let a and b be the first and the last numbers of a G.P., and g1 , g2 , ...... gn are n geometric means between a and b. Then a, g1 , g2 ...... gn, b, are in G.P. The number of means = n The total number of terms in the G.P. (N) = n + 2 Then, last term (t n) = arN – 1 or, b = ar(n + 2) – 1 or, b a = rn + 1 ∴ r = 1 b n+1 a Again, g1 = t2 = ar, g2 = t3 = ar2 , g3 = t4 = ar3 , …, gn = arn 7.9 Relation between arithmetic mean and geometric mean The arithmetic mean (A.M.) is always greater than the geometric mean (G.M.) between two unequal positive numbers. Let a and b be two unequal positive numbers. Then, A.M. between a and b = a + b 2 and G.M. between a and b = ab Now, A.M. – G.M. = a + b 2 – ab = a + b – 2 ab 2 = ( a) 2 – 2. a. b + ( b) 2 2 = ( a – b) 2 2 = Positive quantity Also, A.M. – G.M. > Positive quantity or, A.M.=G.M. + positive quantity ∴ A.M. > G.M. Hence, the geometric mean is greater than the geometric mean between two unequal positive numbers. Facts to remember 1. G.M. between any two numbers a and b = ab 2. If there are n geometric means between any two numbers a and b then common ratio is obtained by the formula, r = 1 b n+1 a 3. 1st mean (g1 ) = 2nd term = ar, 2nd mean (g2 ) = 3rd term = ar2 , ……. 4. Last mean or nth mean (gn) = arn or, b r 5. If A.M. is the arithmetic mean and G.M. is the geometric mean between two unequal positive numbers, then A.M. > G.M.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 149 Vedanta Excel in Mathematics - Book 10 Sequence and Series Worked-out Examples Example 1: Find the geometric mean between 2 and 18. Solution: Here, the first term (a) = 2 and the last term (b) = 18 Now, G.M. = ab = 2 × 18 = 36 = 6 Hence, the geometric mean between 2 and 18 is 6. Example 2: The geometric mean between x and 8 is 4, find the value of x. Also find the arithmetic mean between them. Solution: Here, the first term (a) = x, the last term (b) = 8 and G.M. = 4 Now, G.M. = ab or, 4 = x × 8 On squaring both sides, we get 16 = 8x ∴ x = 2 So, the first term (a) = x = 2 Again, A.M. = a + b 2 = 2 + 8 2 = 5 Hence, the required arithmetic mean is 5. Example 3: If A.M. is the arithmetic mean and G.M. is the geometric mean between 20 and 45. Show that A.M. > G.M. Solution: Here, the first term (a) = 20 and the last term (b) = 45 Now, A.M. = a + b 2 = 20 + 45 2 = = 32.5 Also, G.M. = ab = 20 × 45 = 900 = 30 Hence, A.M. > G.M. Example 4: Insert 4 geometric means between 2 3 and 162. Solution: Here, the first term (a) = 2 3 , the last term (b) = 162 and number of geometric means (n) = 4 Now, common ratio (r) = 1 b n+1 a = 1 162 4+1 2/3 = (243) 1 5 = (3)5 × 1 5 = 3 Alternative process for finding r: No. of terms (n) = 4 + 2 = 6 Now,sixth term (t6 ) = 162 or, arn – 1 = 162 or, 2 3 r6 – 1 = 162 or, r5 = 243 = 35 ∴ r = 3

Vedanta Excel in Mathematics - Book 10 150 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Again, g1 = ar = 2 3 × 3 = 2, g2 = ar2 = 2 3 × 32 = 6 g3 = ar3 = 2 3 × 33 = 18, g4 = ar4 = 2 3 × 34 = 54 Hence, 2, 6, 18 and 54 are the required 4 geometric means between 2 3 and 162. Example 5: There are n geometric means between 3 and 96. If the third mean is 24, find: (i) the common ratio (ii) the value of n (iii) the remaining means Solution: Let g1 , g2 , g3 , …, gn be the n geometric means between 3 and 96 respectively. Then, 3, g1 , g2 , 24, …, gn, 96 are in a G.P. Here, the first term (a) = 3, the last term (b) = 96 and the third mean (m3 ) = 24 (i) Now,the third mean (g3 ) = ar3 or, 24 = 3r3 or, 8 = r3 or, r = 2 Hence, the common ratio is 2. (ii) Also, r = 1 b n+1 a or, 2 = 1 96 n+1 3 or, 2 = (32) 1 n+1 or, 2 = (2)5 × 1 n+1 or, 1 = 5 n + 1 or, n + 1 = 5 or, n = 4 Hence, there are 4 geometric means between 3 and 96. Example 6: There are n geometric means between 5 and 80. If the ratio of the first mean to the last mean is 1: 4, find: (i) the common ratio (ii) the value of n (iii) the means Solution: Let g1 , g2 , g3 , …, gn be the n geometric means between 5 and 80 respectively. Then, 5, g1 , g2 , , …, gn, 80 are in a G.P. Here, the first term (a) = 5, the last term (b) = 80 Now, the first mean (g1 )= ar = 5r and the last mean (gn) = b r = 80 r By question, g1 gn = 1 4 or, 5r 80/r = 1 4 Checking Third mean (g3 )= ar3 = 3 × 23 = 24, which is given in the question. (iii) Again, g1 = ar = 3 × 2 = 6, g2 = ar2 = 3 × 22 = 12 g4 = ar4 = 3 × 24 = 48 Hence, the remaining means are 6, 12 and 48.