147 Check Your Performance 4. (i) The height of a cylindrical tin can of the base circumference 42 cm is 24 cm. (ii) The diameter of the base of a cylinder log of the height 20 cm is 28 cm. (a) Find the radius of the circular base of the can. (b) Find the curved surface area of the can. (c) Find the total surface area of the can. (d) Find the volume of the can. Solution: (i) Here, height of a cylindrical tin (h) = 24 cm circumference (c) = 42 cm (a) radius (r) = ? Now, C = 2r or, 42 = 2 × 22 7 × r r = 42 × 7 2 × 22 = 6.68 cm (b) CSA = ? Now, CSA = 2rh = c × h = 42 × 24 = 1008 cm2 (c) TSA = 2r(r + h) = 42(6.68 + 24) = 1288.56 cm2 (d) Volume (V) = r 2 h = 22 7 × (6.68)2 × 24 = 3365.80 cm3 (ii) Here, diameter of the base (d) = 28 cm (a) radius (r) = d 2 = 28 2 = 14 cm height (h) = 20 cm (b) CSA = 2rh = 2 × 22 7 × 14 × 20 = 1760 cm2 (c) TSA = 2r(r + h) = 2 × 22 7 × 14(14 + 20) = 2992 cm2 (d) Volume (V) = r 2 h = 22 7 × 142 × 20 = 12,320 cm3 5. (a) If the lateral surface of a right circular cylinder is 3960 cm2 and its base radius is 10 cm, find its height. (b) If the volume of a right circular cylinder of height 28 cm is 13200 cm3 , find the radius of the base. (c) The circumference of the base of a cylinder is 33 cm and the area of its curved surface is 528 cm2 . Find its height. (d) The volume of a cylinder is 3080 cu. cm and its diameter is 14 cm. Find its height. Solution: (a) Here, LSA of right circular cylinder = 3960 cm2 radius (r) = 10 cm height (h) = ?
( ( ( 6. O ( ( ( Now, LSA = or, 2rh = or, 2 × 22 7 or, h = 39 2 × (b) Here, volum height radius Now, V = 13 or, r 2 h = or, 22 7 × r2 or, r2 = 13 2 or, r2 = 15 r = 1 (c) Here, circum CSA = height Now, CSA = or, 33 × h h = 528 33 (d) Here, volum diamet radius height Now, V = r or, 3080 = h = 30 22 Observe the follo (a) Calculate th (b) Calculate th (c) Calculate th = 3960 = 3960 × 10 × h = 3960 960 × 7 × 22 × 10 = 126 c me of cylinder (V) (h) = 28 cm (r) = ? 3200 13200 2 × 28 = 13200 200 × 7 22 × 28 0 50 = 12.25 cm mference (c) = 33 = 528 cm2 (h) = ? = 528 = 528 8 3 = 16 cm me (V) = 3080 cm ter (d) = 14 cm (r) = d 2 = 14 2 = 7 c (h) = ? r 2 h = 22 7 × 72 × h 080 × 7 = 20 cm owing objects an he curved surfac he total surface he volume of the 148 cm ) = 13200 cm3 3 cm m3 cm nd answer the gi ce area of the giv area of the given e given solids. [ c = 2r] iven questions. ven solids. n solids.
149 Solution: (i) Here, radius (r) = 10.5 cm (a) CSA = 2rh = 2 × 22 7 × 10.5 × 24 = 1584 cm2 (b) TSA = 2r(r + h) = 2 × 22 7 × 10.5 (10.5 + 24) = 2277 cm2 (c) Volume (V) = r 2 h = 22 7 × (10.5)2 × 24 = 8316 cm3 (ii) Here, diameter (d) = 14 cm radius (r) = d 2 = 14 2 = 7 cm height (h) = 30 cm (a) CSA = 2rh = 2 × 22 7 × 7 × 30 = 1320 cm2 (b) TSA = 2r(r + h) = 2 × 22 7 × 7(7 + 30) = 1628 cm2 (c) Volume (V) = r 2 h = 22 7 × 72 × 30 = 4620 cm3 (iii) Here, diameter (d) = 21 cm radius (r) = d 2 = 21 2 = 10.5 cm height (h) = 40 cm (a) CSA of hemi-cylinder = rh = 22 7 × 10.5 × 70 = 1320 cm2 (b) TSA of hemi-cylinder = r(r + h) + 2rh = 22 7 × 10.5(10.5 + 40) + 2× 10.5 × 40 = 1666.50 + 840 = 2506.50 cm2 (c) Volume of hemi-cylinder (V) = 1 2 r 2 h = 1 2 × 22 7 × (10.5)2 × 40 = 6930 cm3 (iv) Here, radius (r) = 6 cm height (h) = 20 cm (a) CSA given solid = 2rh × 3 4 = 3 2 rh = 3 2 × 22 7 × 6 × 20 = 377.24 cm2 (b) TSA = 3 2 rh + 2rh = 377.14 cm2 + 2 × 6 × 20 = 377.14 cm2 + 240 cm2 = 617.14 cm2 (c) Volume (V) = 3 4 × r 2 h = 3 4 × 22 7 × 62 × 20 = 1697.14 cm3 .
7. ( ( Soluti ( ( 8. ( ( ( Soluti ( ( (i) Find the qu cylindrical (ii) The cross s radius 4 me (a) Calcul (b) How m ion: (i) Here, diame radius height Now, capaci Now, quanti (ii) Here, radius length (a) Area o (b) Quanti (i) Find the re brisket fire inches as sh (ii) The thickne (a) Find th (a) Find th (iii) A tin of bea height of 10 (a) What i (b) What ion: (i) Here, radius height volume volume The req (ii) Here, thickn radius uantity of wate drum. section of a 240 eters. late the area of t much air can con eter (d) = 60 cm (r) = d 2 = 60 2 = 30 (h) = 130 cm ity of drum (V) ity of water = 1 2 × s of semi-circle (r of tunnel (h) = 2 f cross section (A ity of air in tunne equirement of r e woods with r hown in the adjo ess of the CD is 0 he volume of on he volume of suc ans is cylindrica 0 cm. is its capacity? area of lid is req s of fire woods (r) (h) = 3 inches e of 1 fire woods e of 14 fire wood quirement of raw ness of CD (h) = 0 (r) = 6 cm 150 er in liter conta 0 m long tunnel the cross section ntain in the tunn 0 cm = r 2 h = 22 7 × = 367.71 litre × 367.71 litre = 1 r) = 4 m 40 m A) = area of sem = r 2 2 = 22 7 × 4 2 el = V = r 2 h 2 = 22 7 raw materials to radius 5 inches oining figure (ne 0.15 cm and rad e CD. ch 24 CDs. l shape whose b quired to cover i ) = 5 inches = r 2 h = 22 7 × 52 ds = 14 × 235.71 w material to mak 0.15 cm ining in the giv l is a semicircle n of the tunnel, nel ? Calculate it 302 × 130 = 367 e [ 1000 cm3 = 83.85 litre. mi-circle 42 = 25.14 m2 2 7 × 42 × 240 2 = 60 o make the 14 and height 3 eglect pores). dius 6 cm. base has radius o it ? 2 × 3 = 235.71 cub = 3299.94 cubic ke the 14 brisket w ven of t. 7714.29 cm3 = 1 litre] 034.28 m3 of 4 cm and bic inches inches wood = 3299.94 cubic inches
( 9. ( ( ( Soluti ( ( (a) The vo (b) The vo (iii) Here, radius height (a) Capaci (b) Area o (i) The cylindr (a) Calcul (b) Find t fill it. (ii) Observe the tanker. (a) Calcul (b) Find t fill it. (iii) A manufact is 6 mm in d (a) Find th (b) What v ion: (i) Here, height radius (a) Capaci (b) No. of (ii) Here, by the radius height (a) Capaci (b) The no olume of CD (V) olume of such 24 s (r) = 4 cm (h) = 10 cm ity of it (V) = r 2 f lid to cover it (C rical tea urn is 9 late capacity in l he number of 2 e given cylindric late the capacity he number of 2 turer is making diameter and 7 c he volume of eac volume of plasti t of tea urn (h) = (r) = 30 cm ity in litre (V) = = = 200 m cups nee e given oil-chamb (r) = 1.5 2 = 0.75 m (h) = 4.5 m ity of the tanker ( o. of 200 drum n 151 = r 2 h = 22 7 × 62 CD's = 24 × 16.9 2 h = 22 7 × 42 × 10 CSA) = 2rh = 2 0 cm high, and h litres. 200 ml cups need cal oil-chamber y of the tanker. 00 drums need a batch of 1000 cm long. ch needle. ic will be needed 90 cm = r 2 h = 22 7 × 302 = 254571.43 1000 litre = 254.57 litres eded to fill it ber diameter (d) = m (V) = r 2 h = 2 7 = 7.96 × 1 needed to fill it × 0.15 cm = 16.9 97 cm3 = 407.28 = 502.86 cm3 × 22 7 × 4 × 10 = has a base of rad ded to of the ded to 00 plastic cylind d altogether? 2 × 90 cm3 = 254 es [ 1 cm3 = 10 = 254.57 litres 200 m = = 1272.85 = 1272 (Approx = 1.5 m 22 7 × (0.75)2 × 4.5 1000 litre = 7960 = 7960 litres 200 litres = 3 = 39 (Approxim 97 cm3 cm3 251.43 cm2 dius 30 cm. rical knitting ne 571.43 cm3 000 litre] = 254.57 × 1000 m 200 m ximately) 5 = 7.96 m3 litre. 39.8 mately) eedles. Each one m
152 (iii) Here, diameter of each knitting (d) = 6 mm radius = 6 2 = 3 mm = 0.3 cm length of knitting (h) = 7 cm (a) Now, volume of 1 needle (V) = r 2 h = 22 7 × (0.3)2 × 7 = 1.98 cm3 (b) The volume of plastic needles (V) = 10,000 × 1.98 m3 = 19,800 cm3 . 10. (a) The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm3 . Find its radius. (b) The radii of two cylinders are in the ratio 4 : 3. Find the ratio of their volumes if the heights are taken to be the same. (c) The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 4. Calculate the ratio of their curved surface areas. (d) The radius of the base of a right circular cylinder is halved. What is the ratio of the volume of the reduced cylinder to that of the original one? Solution: (a) Here, the ratio of radius and height of a cylinder = 5 : 7 So, let the radius be 5x and height be 7x. Volume (V) = 550 cm3 Now, V = r 2 h or, 550 = 22 7 × (5x)2 × 7x or, 550 = 22 × 25x3 or, 550 22 × 25 = x3 or, 1 = x3 x = 3 1 = 1 The radius (r) = 5x = 5 × 1 = 5 cm. (b) Let the radii of 1st cylinder be r and 2nd be R. then, r : R = 4 : 3 Also, the ratio of their volumes = ? Now, if the v be the volume of 1st and V be the volume of 2nd cylinder. then, v V = r 2 h R2 h or, v V = 42 32 or, v V = 16 9 v : V = 16 : 9 (c) Here, let r and R be two radii of two cylinder then, r R = 2 3 Also, let h and H be the heights of two cylinder resp. then, h H = 5 4 Also, let the CSA of two cylinders be A1 and A2 respectively. Then, A1 A2 = 2rh 2RH = 2 3 × 5 4 = 5 6 Hence, the required ratio of their curved surface is 5 : 6.
153 (d) Here, let the radius of base of cylinder be r. then it's volume (V1) = r 2 h Again, when the radius of base in half, then it's volume (V2) = 1 r 2 × h = r 2 h 4 Now, ratio of reduced cylinder to that of the original = V2 V1 = r 2 h 4 r 2 h = r 2 h 4 × 1 r 2 h = 1 4 = 1 : 4. 11. (i) A solid cylinder has the total surface area of 462 sq. cm. Its curved surface area is onethird of its total surface area. (ii) A solid cylinder has a total surface area of 231 cm2 . Its curved surface area is 2 3 rd of the total surface area. (a) Find the height of the solid. (b) Find the base radius of the solid. (c) Find the volume of the cylinder. Solution: (i) Here, TSA of solid cylinder = 462 cm2 CSA = 1 3 of TSA (a) height of cylinder = ? Now, CSA = 1 3 of TSA or, 2rh = 1 3 × 462 cm2 or, 2rh = 154 cm2 Again, TSA = 462 or, 2rh + 2r 2 = 462 or, 154 + 2r 2 = 462 or, 2r 2 = 462 – 154 or, 2 × 22 7 × r2 = 308 or, r2 = 308 × 7 44 = 49 r = 49 = 7 cm Again, 2rh = 154 or, 2 × 22 7 × 7 × h = 154 h = 154 44 = 3.5 cm. Height of cylinder = 3.5 cm. (b) The base radius (r) = 7 cm (c) The volume (V) = r 2 h = 22 7 × 7 × 3.5 = 539 cm3
154 (ii) Here, TSA = 231 cm2 CSA = 2 3 of TSA = 2 3 × 231 CSA = 154 cm2 (a) For the height of the solid TSA = 2rh + 2r 2 or, 231 = 154 + 2r 2 or, 231 – 154 = 2 × 22 7 × r2 or, 77 × 7 44 = r2 or, 539 44 = r2 or, 12.25 = r2 r = 12.25 = 3.5 cm Again, CSA = 154 or, 2rh = 154 or, 2 × 22 7 × 3.5 × h = 154 h = 154 × 7 2 × 22 × 3.5 = 7 cm Hence, the height of cylinder = 7 cm. (b) The base radius (r) = 3.5 cm (c) Volume of the cylinder (V) = r 2 h = 22 7 × (3.5)2 × 7 = 22 × 12.25 = 269.5 cm3 12. (i) A rectangular sheet of aluminium foil is 44 cm long and 20 cm wide. A cylinder is made out of it by rolling the foil along its wide. (a) Find the volume of the cylinder. (b) Calculate the cost of painting on its outer and inner surfaces at Rs. 1.25/cm2 . (ii) A rectangular sheet of a 44 cm long and 20 cm broad metal is rolled along its length into a cylinder so that the cylinder has 20 cm as its height. (a) Calculate the cost of iron to use in it at Rs. 18.50 per sq. cm. (b) If its outer surface is painted, how much cost for it at Rs. 1.25 per sq. cm will heed? Solution: (i) Here, length of aluminium foil () = 44 cm wide (b) = 20 cm A cylinder is made out of it by rolling the foil along its wide (a) Volume of the cylinder (V) = ? by question, circumference (c) = 20 cm or, 2r = 20 cm or, 2 × 22 7 × r = 20 cm r = 20 × 7 44 = 3.18 cm and height of the cylinder (h) = 44 cm Now, V = r 2 h = 22 7 × (3.18)2 × 44 = 1398.40 cm3
155 20 cm 2 cm 35 cm (b) The cost of painting on it's outer and inner surfaces = ? For then, we have area of CSA = 2 × 2rh = 4 × 22 7 × 3.18 × 44 = 1758.99 cm2 Now, total cost at the Rs. 1.25 = Rs. 1758.99 × 1.25 = Rs. 1298.74. (ii) Here, when the metal is rolled along it's length 44 cm becomes the circumference of the base of the cylinder and broad of the metal 20 cm becomes it's height. So, C = 44 cm or, 2r = 44 cm or, 2 × 22 7 × r = 44 cm r = 44 × 7 44 = 7 cm and height (h) = 20 cm (a) volume of iron used in it (V) = r 2 h = 22 7 × 72 × 20 = 22 × 7 × 20 = 3080 cm3 Since, cost of 1 cm2 iron = Rs. 18.50 The cost of iron used in it = Rs. 3080 × 18.50 = Rs. 56,980. (b) Surface area of outer (CSA) = 2rh = 2 × 22 7 × 7 × 20 = 880 cm2 Cost of 1 cm2 panting = Rs. 1.25 The cost of panting = Rs. 880 × 1.25 = Rs. 1100 13. (i) A cylindrical hollow iron pipe has an external diameter 20 cm. It is 35 cm long and thickness of iron is 2 cm. (a) Find the volume of the metal used. (b) Calculate the cost of iron at Rs. 5.50 per cu. cm. (ii) A cylindrical road roller made of iron, is 1 m long. Its inner diameter is 54 cm and the thickness of the iron sheet rolled into the road roller is 9 cm. (a) Find the weight of the roller if 1 c.c. of iron weights 8 gm. (b) Calculate the cost of iron at Rs. 5.50 per cu. cm. Solution: (i) Here, external diameter (d) = 20 cm radius (R) = d 2 = 20 2 = 10 cm internal diameter = 20 cm – 2 – 2 = 16 cm internal radius (r) = 16 2 = 8 cm height of cylinder hollow iron pipe (h) = 35 cm thickness of iron = 2 cm (a) the volume of the metal used = R2 h – r 2 h = h(R2 – r2 ) = 22 7 × 35(102 – 82 ) = 110(100 – 64) = 110 × 36 = 3960 cm3
156 (b) Rate of iron per cm3 = Rs. 5.50 the total cost of iron = Rs. 3960 × 5.50 = Rs. 21280 (ii) Here, length of cylindrical road roller (h) = 1 m = 100 cm inner diameter (d) = 54 cm inner diameter (r) = 54 2 = 27 cm thickness of the iron = 9 cm Now, outer diameter (d) = 54 cm + 9 cm + 9 cm = 72 cm outer radius (R) = 72 2 = 36 cm Now volume of iron used in it (V) = R2 h – r 2 h = h(R2 – r2 ) = 22 7 × 100(362 – 272 ) = 2200 7 (1296 – 729) = 2200 7 × 567 = 178,200 cm3 (a) weight of 1 cm3 iron = 8 gm weight of 178,200 cm3 iron = 178,200 × 8 gm = 1425,600 gm = 1425.6 kg (b) Rate of cost of 1 cm3 iron = Rs. 5.50 The cost of iron used = Rs. 178200 × 5.50 = Rs. 980,100.
CHA 7.1 Rea Keep 1. ( ( ( ( ( ( ( ( ( ( Soluti ( ( ( ( ( ( ( ( ( (j Chec 2. O APTER 7 Surface Ar ad / Underst ping Skill Sha (a) Define sphe (b) Define hem (c) What is the (d) Write down (e) Write down radius is giv (f) What is the (g) What is the (h) Write the c (i) What is the (j) What is the ion: (a) A round so called spher (b) A half of a s (c) The area of (d) The surface (e) The curved (f) The total sur (g) The volume (h) The curved (i) The surface (j) The radius o ck Your Perfo Observe the follo Radius = SPHERE rea and Vol tand / Think arp ere. misphere. e area of great ci n the formula to n the formula t ven. formula to calcul e volume of a sph urved surface ar e surface area of e radius of a hem lid shape in wh re. sphere is called a great circle of a s area of a sphere surface area of a rface area of a he of a sphere havin surface area of a area of a hemisp of a hemisphere w ormance owing solids and = 12.1 cm R 157 lume of Sph PRACTICE / Do ircle of a sphere compute the su o calculate the late the total surfa here having radi rea of a hemisph f a hemisphere h misphere with cu hich every point hemisphere. sphere having rad with radius r is 4 hemisphere whe emisphere having ng radius 21 cm sphere having ra phere having radiu with curved surfa d answer the foll Radius = 12.5 cm here E 7.1 having radius r rface area of a s curved surface face area of a hem ius 21 cm? here having radi having radius 3.5 urved surface ar on its surface eq dius r is r 2 . 4r 2 . en its radius is giv g radius r is 3r 2 . is 38808 cm2 . adius 14 cm is 12 us 3.5 cm is 115. ace area 8 square lowing questions m Diameter = r ? sphere with radi area of a hemi misphere having r ius 14 cm. 5 cm? ea 8 square unit quidistant from i ven, is 2r 2 . 32 cm2 . 50 cm2 . e units is 2 units. s. = 7 cm Dia ius r. sphere when its radius ‘r’ units? ts? its fixed point is ameter = 10 cm s s
158 (a) Write the definition of sphere. (b) Find the surface area of the above spherical solids. (c) Find the volume of the above spherical solids. (d) Find the relation between the total surface area and volume of each spherical solid. Solution: (i) Here, radius (r) = 12.1 cm (b) The surface area of sphere = 4r 2 = 4 × 22 7 × (12.1)2 = 1840.58 cm2 (c) Volume of the sphere (V) = 4 3 r 3 = 4 3 × 22 7 × (12.1)3 = 7423.68 cm3 (d) The relation between TSA and the volume of sphere is TSA V = 4r 2 4 3 r3 or, TSA V = 3 12.1 or, 3V = 12.1 TSA (ii) Here, radius (r) = 12.5 cm (b) Surface area of sphere = 4r 2 = 4 × 22 7 × (12.5)2 = 1964.29 cm2 (c) Volume of sphere (V) = 4 3 r 3 = 4 3 × 22 7 × (12.5)3 = 8184.52 cm3 (d) The relation between TSA and volume (V) is TSA V = 4r 2 4 3 r3 or, TSA V = 3 r or, TSA V = 3 12.5 3V = 12.5 TSA. (iii) Here, diameter (d) = 7 cm radius (r) = d 2 = 7 2 = 3.5 cm (b) Surface area (A) = 4r 2 = 4 × 22 7 × (3.5)2 = 154 cm2 (c) Volume (V) = 4 3 r 3 = 4 3 × 22 7 × (3.5)3 = 179.67 cm3 (d) Relation between surface area and volume is A V = 4r 2 4 3 r3 or, A V = 3 r or, A V = 3 3.5 3V = 3.5 A.
( 3. O ( ( ( ( Soluti ( ( (iv) Here, diame radius (b) Surface (c) Volum (d) Relatio A V or, A V 3V Observe the follo (a) Write the fo (b) Find the cu (c) Find the vo (d) Write the r ion: (i) Here, radius (a) The for (b) CSA o Also, T (c) Volum (d) The rat (ii) Here, diame radius (b) CSA o Also, T eter (d) = 10 cm. (r) = 10 2 = 5 cm e area (A) = 4r 2 me (V) = 4 3 r 3 = 4 3 on between surfac A V = 4r 2 4 3 r3 A V = 3 r V = 5A. owing solids and ormula to calcul urved surface are olume of the abov atio of the curve s (r) = 14 cm rmula to calculat f given hemisphe TSA of hemisphe me of given hemis tio of CSA and T eter (d) = 3.5 ft (r) = 3.5 2 = 1.75 f f hemisphere = 2 TSA of hemisphe 159 2 = 4 × 22 7 × (5)2 4 3 × 22 7 × (5)3 = 52 ce area and volum d answer the foll late the volume ea and the total ve hemispheres. ed surface area a te the volume of a ere = 2r 2 = 2 × 2 ere = 3r 2 = 3 × 2 7 sphere (V) = 2 3 r TSA = CSA TSA = 123 184 ft 2r 2 = 2 × 22 7 × (1 ere = 3r 2 = 3 × 2 7 = 314.29 cm2 23.81 cm3 me is or, A V = 3 5 lowing questions of a sphere. surface area of . and total surfac a sphere = 4 3 r 3 . 22 7 × (14 cm)2 = 1 22 7 × (14 cm)2 = 1 3 = 2 3 × 22 7 × (14 c 32 cm2 48 cm2 = 154 231 = 1 1.75 ft)2 = 19.25 f 22 7 × (1.75 ft)2 = 2 s. the above hemis e area of the hem 1232 cm2 848 cm2 cm)3 = 5749.33 c 54 : 231 = 2 : 3 ft2 28.86 ft2 spheres: misphere. cm3
( ( 4. ( ( ( ( (c) Volum (d) The rat (iii) Here, circum or, 2r = 4 r = 44 4 (b) CSA o Also, T (c) Volum (d) The rat (iv) Here, diame radius (b) CSA o Also, T (c) Volum (d) The rat (i) A sphere ha (a) Write curved (b) Find th (c) Find th (ii) A spherical (a) Write (b) Find th (c) Find th (iii) A sphere ha (a) Write (b) What i (c) Find th (iv) The volume (a) Write (b) Find th (c) How m me of hemisphere tio of CSA and T mference (c) = 44 44 cm × 7 4 = 7 cm f hemisphere = 2 TSA of hemisphe me of hemisphere tio of CSA and T eter (d) = 12742 k (r) = 12742 2 = 63 f hemisphere = 2 TSA of hemisphe me of hemisphere tio of CSA and T as a circumferen the relation bet d surface area of he radius of the he surface area l iron ball has di the relation betw he radius of the he surface area as the surface ar the formula to c is the radius of t he volume of the e of a sphere is 3 the formula to c he diameter of it many outer layer 160 (V) = 2 3 r 3 = 2 3 × TSA = CSA TSA = 19. 28. 4 cm 2r 2 = 2 × 22 7 × (7 ere = 3r 2 = 3 × 2 7 (V) = 2 3 r 3 = 2 3 × TSA = CSA TSA = 308 462 km 71 km 2r 2 = 2 × 22 7 × (6 ere = 3r 2 = 3 × 2 7 (V) = 2 3 r 3 = 2 3 × TSA = CSA TSA = 255 382 nce of the great c tween the surfa f a hemi-sphere. sphere. and volume of t iameter 7 cm. ween the volume ball. and volume of t rea 154 cm2 . calculate the vol the given sphere e sphere. 38808 cm3 . calculate the tota ts great circle. r of the sphere a × 22 7 × (1.75 ft)3 = .25 .86 = 2 : 3 or, 2 × 22 7 × r 7 cm)2 = 308 cm2 22 7 × (7 cm)2 = 46 × 22 7 × (7 cm)3 = 7 8 2 = 2 3 = 2 : 3 6371 km)2 = 2551 22 7 × (6371 km)2 = × 22 7 × (6371 km) 5134886.29 2702329.43 = 2 3 = circle 66 cm. ace area of a sp . he sphere. e of a sphere and he sphere. lume of a sphere e. al surface area o are there? Find i = 11.23 ft3 = 44 cm 62 cm2 718.67 cm3 134886.29 km2 = 382702329.43 k ) 3 = 5.418214535 = 2 : 3 here and the d its hemi-spher e. of a sphere. it. km2 × 1011 km3 re.
161 Solution: (i) Here, circumference of great circle of sphere (c) = 66 cm (a) the relation between the surface are of sphere and the curved surface area of a hemisphere TSA CSA = 4r 2 2r 2 = 2 or, TSA = 2 CSA CSA of hemisphere is half of TSA of sphere. (b) Here, c = 66 cm or, 2r = 66 cm or, 2 × 22 7 × r = 66 cm r = 66 × 7 44 = 10.5 cm (c) The surface area = 4r 2 = 4 × 22 7 × (10.5)2 = 1386 cm2 and volume (V) = 4 3 r 3 = 4 3 × 22 7 × (10.5)3 = 4851 cm3 (ii) Here, diameter of spherical ball (d) = 7 cm (a) Volume of sphere = 2 × the volume of hemisphere. (b) diameter (d) = 7 cm radius (r) = d 2 = 7 cm 2 = 3.5 cm (c) The surface area = 4r 2 = 4 × 22 7 × (3.5 cm)2 = 154 cm2 and volume of sphere (V) = 4 3 r 3 = 4 3 × 22 7 × (3.5 cm)3 = 179.67 cm3 (iii) Here, the surface area (TSA) = 154 cm2 (a) Volume of the sphere (V) = 4 3 r 3 (b) Now, TSA = 154 cm2 or, 4r 2 = 154 cm2 or, 4 × 22 7 × r2 = 154 cm2 or, r2 = 154 × 7 88 cm2 = 12.25 cm2 r = 12.25 = 3.5 cm (c) Volume (V) = 4 3 r 3 = 4 3 × 22 7 × (3.5 cm)3 = 179.67 cm3 (iv) Here, volume of sphere (V) = 38808 cm3 (a) The formula to find the total surface area = 4r 2 (b) Now, V = 38808 or, 4 3 r 3 = 38808 or, 4 3 × 22 7 × r3 = 38808 or, r3 = 38808 × 21 88 = 9261
162 r = 3 9261 = 21 cm Diameter (d) = 2r = 2 × 21 = 42 cm (c) Outer layer of the sphere = 4r 2 = 4 × 22 7 × (21 cm)2 = 5544 cm2 5. (a) Find the radius of a sphere whose surface area is equal to the area of a circle with diameter 5.6 cm. (b) Find the volume of a sphere whose area of the great circle is equal to the curved surface area of a hemisphere with circumference of great circle 44 cm. (c) Calculate the volume of a sphere whose surface area is 36π m2 . Solution: (a) Here, diameter of a circle (d) = 5.6 cm radius (r) = 5.6 2 = 2.8 cm Now, area of circle = r 2 = 22 7 × (2.8 cm)2 = 24.64 cm2 By question, the surface area of sphere = area of circle or, 4r 2 = 24.64 or, 4 × 22 7 × r2 = 24.64 or, r2 = 24.64 × 7 88 or, r2 = 1.96 r = 1.96 = 1.4 cm (b) Here, circumference of great circle (c) = 44 cm or, 2r = 44 cm or, r = 44 2 = 44 2 × 22 7 cm r = 7 cm Now, area of curved surface of hemisphere = 2r 2 = 2 × 22 7 × 72 = 308 cm2 By question, the area of great circle = 308 cm2 or, r 2 = 308 cm2 or, 22 7 × r2 = 308 cm2 or, r2 = 308 × 7 22 cm2 = 98 cm2 r = 98 cm2 = 9.89 cm The volume of the sphere = 4 3 r 3 = 4 3 × 22 7 × (9.89 cm)3 = 4053.71 cm3 (c) The surface area of sphere = 36 m2 or, 4r 2 = 36 or, r2 = 36 4 or, r2 = 9 r = 9 = 3 m The volume of the sphere = 4 3 r 3 = 4 3 × 22 7 × 33 = 113.14 cm3
6. ( ( ( ( Soluti ( ( (i) A manufact sphere of th (a) How m (b) What i (c) How m (d) If it de balls w (ii) How many moulded fr cm and diam (a) Calcul (b) Calcul (c) Find th (iii) How many radius, each (iv) How many rectangular ion: (i) Here, radius volume Again, diam Volum Now, no. of (b) The vo (c) He can (d) Radius Now, v N Hence, (ii) Here, diame radius (a) Now, v (b) Again, turer company he diameter 16 c much lead does i is the volume of many small lead ecreases 10% on will form ? solid spheres ea om a solid meta meter 4 cm with late the volume o late the volume o he number of sp y spherical bulle h bullet being 12 y spherical lead r solid of lead wi s of small lead ba e of one lead ball meter of a sphere ( me of a sphere (V2 f lead balls (N) = olume of each sm n form 512 small s of lead ball after volume of lead ba No. of balls = 88 88 21 × , we have 702 – 5 eter of solid spher (r) = 6 cm 2 = 3 cm volume (V1) = 4 3 height of cylinde 163 wants to make t cm. it have ? Find it. f each small lead balls can he for n the radius of t ach 6 cm in diam al cylinder whose hout any loss? of the sphere. of the cylinder. phere formed fro ets can be mad 2 mm in diamete d shots each 4 ith dimensions 6 alls (r) = 1 cm ls (V1) = 4 3 r 3 = 4 3 (d) = 16 cm 2) = 4 3 r 3 = 4 3 × 2 7 V2 V1 = 2145.52 cm 4.19 cm3 mall lead ball = 88 21 lead balls from th r decreasing 10% all (V3) = 4 3 × 22 7 × 512 21 × 0.729 = 702 512 = 190 more b re (d) = 6 cm m r 3 = 4 3 × 33 = 3 er (h) = 45 cm the small lead b . d ball ? Find it. m from the lead the small lead b meter can be e height is 45 om the cylinder. de from a cylind er? .2 cm in diam 66 cm, 42 cm and 4 3 × 22 7 × 13 = 88 21 2 7 × 83 = 88 × 512 21 m3 = 512.05 ~ 512 8 1 cm3 = 4.19 cm3 he lead of sphere % = 1 cm – 10% o × (0.9)3 = 88 21 × 0 balls. 36 cm3 balls each of rad d of sphere ? Fin ball, how many m . der 90 mm lon meter can be ob d 21 cm? cm3 = 4.19 cm3 2 cm3 = 2145.52 2 e. of 1 = 1 cm – 10 100 0.729 dius 1 cm from a nd it. more small lead g and 4 mm in btained from a cm3 0 × 1 = 0.9 cm a d n a (a)
164 diameter (d) = 4 cm radius = 4 2 = 2 cm Now, volume of cylinder (V2) = r 2 h = × 22 × 45 = 180 cm3 (c) No. of spheres formed from the cylinder = V2 V1 = 180 36 = 5 (iii) Here, diameter of bullet (d) = 12 mm radius (r) = 12 2 = 6 mm Now, volume of bullet (V1) = 4 3 r 3 = 4 3 × 63 = 288 Again, radius of cylinder (r) = 4 mm height (h) = 90 mm Now, volume of cylinder (V2) = r 2 h = × 42 × 90 = 1440 No. of bullets (spherical) = V2 V1 = 1440 288 = 5 (iv) Here, diameter of spherical lead (d) = 4.2 cm radius = 4.2 2 = 2.1 cm. Volume (V1) = 4 3 r 3 = 4 3 × (2.1)3 = 4 3 × 22 7 × 9.261 = 33.808 cm3 Again, dimension of rectangular solid = 66 cm × 42 cm × 21 cm Volume of it (V2) = 58,212 cm3 Now, no. of leads can be made = V2 V1 = 58212 cm3 38.808 cm3 = 1500 8. (i) A solid right circular cylinder has a base radius of 12 cm and height of 16 cm. It is melted and made into eight spherical balls of equal size. (a) Find the volume of the cylinder and spherical ball. (b) Find the radius of the spherical ball. (c) Calculate the surface area of each spherical ball. (ii) The diameter of a copper sphere is 6 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 36 cm, find its curved surface area. (iii) The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire. Solution: (i) Here, radius of cylinder (r) = 12 cm height (h) = 16 cm (a) The volume of cylinder (V) = r 2 h = 22 7 × 122 × 16 = 7,241.14 cm3 (b) The cylinder melted and made into 8 spherical balls Volume of cylinder = 8 × volume of sphere or, 7241.14 = 8 × 4 3 r 3
165 or, r3 = 7241.14 × 3 × 7 8 × 4 × 22 = 216 or, r = 3 216 = 6 cm (c) The surface area of each spherical ball = 4r 2 = 4 × 22 7 × (6 cm)2 = 452.57 cm2 (ii) Here, diameter of a sphere (d) = 6 cm Radius (r) = 6 2 = 3 cm Now, volume of it (V) = 4 3 r 3 = 4 3 × × 33 = 4 × × 9 = 36 Again, it is melted and is drawn into a long wire of uniform circular cross-section. length of wire () = 36 cm Now, volume of it = 36 or, r 2 h = 36 or, r2 × 36 = 36 or, r2 = 1 r = 1 cm CSA = 2rh = 2 × 22 7 × 1 × 36 = 226.29 cm2 (iii) Here, diameter of sphere (d) = 6 cm radius (r) = 6 cm 2 = 3 cm The sphere is melted and drawn into a wire whose diameter (d) = 0.2 cm radius (r) = 0.2 2 = 0.1 cm Now, volume of sphere = volume of wire or, 4 3 r 3 = r 2 h or, 4 3 × 33 = (0.1)2 × h or, 36 = 0.1 × h h = 360 cm Hence, the length of the wire = 360 m 9. (i) Three solid spheres of lead are melted into a single solid sphere. If the radii of the three spheres is 1 cm, 6 cm and 8 cm respectively, find the radius of the new sphere. (ii) A sphere of lead of diameter 6 cm is melted and recast into three spherical balls. The diameters of the two of them are 3 cm and 4 cm respectively. Find the diameter of the third. Solution: (i) Here, volume of 1st lead (V1) = 4 3 (1 cm)3 = 4 3 × 1 cm3 Volume of 2nd lead (V2) = 4 3 (6 cm)3 = 4 3 × 216
166 Volume of 3rd lead (V3) = 4 3 (8 cm)3 = 4 3 × 512 Three spheres are melted into a single solid sphere. So, Volume of new sphere = V1 + V2 + V3 or, 4 3 R3 = 4 3 × 1 + 4 3 × 216 + 4 3 × 512 or, 4 3 R3 = 4 3 (1 + 216 + 512) or, R3 = 729 R = 3 729 = 9 cm Hence, radius of new sphere = 9 cm. (ii) Here, diameter of sphere (d) = 6 cm radius (r) = d 2 = 6 2 = 3 cm It's volume (V) = 4 3 r 3 = 4 3 × 33 It is melted into three spherical balls diameter of 1st ball (d1) = 3 cm radius (r1) = 3 2 = 1.5 cm diameter of 2nd ball (d2) = 4 cm radius (r2) = 4 2 = 2 cm Let the radius of 3rd ball, be r Then, volume of lead = V1 + V2 + V3 or, 4 3 × 33 = 4 3 × (1.5)3 + 4 3 × 23 + 4 3 r 3 or, 4 3 × 33 = 4 3 (1.53 + 23 + r3 ) or, 27 = 3.375 + 8 + r3 or, 27 = 11.375 + r3 or, 15.625 = r3 r = 3 15.625 = 2.5 cm The diameter of 3rd ball = 2 × 2.5 cm = 5 cm. 10. (i) A solid metallic sphere of radius 7 cm is cut into two halves. (a) Find the volume of a hemisphere. (b) Find the total surface area of the two hemispheres. (c) What is the ration of the surface area of the sphere and the total surface area of the two hemispheres. (ii) There are two solid hemispherical objects made of earth having the same shape and size. If the total surface area of one hemispherical object is 462cm2 , find the total surface area of the sphere formed by attaching those two hemispherical objects.
Soluti ( ( 11. ( ( ( Soluti ( ion: (i) radius of sol It is cut into (a) The vo (b) Total s (c) Here, T and TS T (ii) Here, TSA o or, 3r 2 = or, 3 × 22 7 or, r2 = 46 or, r2 = 49 r = 7 cm TSA o (i) A spherical cylindrical (a) Are th equal ? (b) Find th (c) If the the wa (ii) A cylinder spherical ir water is rai (iii) A cylindric immersed in ion: (i) Here, diame radius diamet radius (a) Yes, th with a of the v lid metallic spher two halves olume of a hemisp surface area of tw TSA of sphere = SA of two hemisp The ratio of TSA o of one hemispher 462 × r2 = 462 62 × 7 66 9 m f the sphere attac l object of diam vessel of diamet he volume of sp ? Give reason. he volume of a s object is compl ater is raised. of radius 12 cm ron ball is drop ised by 6.75 cm. al jar of radius n the oil. How m eter of spherical o (r) = 14 2 = 7 cm ter of cylindrical (r) = 42 2 = 21 cm he volume of sph specific weight i volume dipped in 167 re (r) = 7 cm phere (V) = 2 3 r 3 wo hemisphere = 2 4r 2 = 4 × 22 7 × 7 phere = 924 cm2 of sphere and TS re = 462 cm2 ching two hemisp meter 14 cm is dr ter 42 cm. pherical object a spherical object. letely immersed m contains wate pped into the c Find the radius 6 cm contains o many spheres are object (d) = 14 cm vessel (d2) = 42 c herical object an is dipped in wate n water so the bot 3 = 2 3 × 22 7 × 73 = 2 × 3r 2 = 6 × 22 7 72 = 616 cm2 SA of two hemisp phere = 4r 2 = 4 × ropped into wat and displaced w . d, find how muc er to a depth of cylinder and th of the ball. oil. Some iron sp e necessary to ra m cm nd displaced wate er the same amou th volumes becom 718.67 cm3 2 × 72 = 924 cm2 phere = 616 cm2 924 cm2 = × 22 7 × 72 = 616 c ter contained in water by it ch level of f 20 cm. A he level of pheres each of ra aise the level of t er by it are equa unt of water gets me equal. = 2 : 3 cm2 . a right circular adius 1.5 cm are the oil by 2 cm? l. When a object displaced which r e t h
168 (b) The volume of spherical object (V) = 4 3 r 3 = 4 3 × 22 7 × 73 = 1437.33 cm3 (c) The object is completely immersed level of water raised (h) = ? Now, volume of spherical object = volume of water displaced or, 4 3 × 73 = r 2 h or, 4 3 × 73 = × (21)2 × h or, 4 3 × 343 = 441 × h h = 4 × 343 3 × 441 = 1.04 cm The level of raised water = 1.04 cm. (ii) Here, radius of cylinder (r) = 12 cm depth of water (h) = 20 cm after spherical iron ball is dropped, level of water raised (h) = 6.75 cm radius of ball (r1) = ? Now, volume of water displaced by ball = r 2 h = 22 7 × 122 × 6.75 = 3054.86 cm3 But, volume of iron ball = volume of water displaced by it or, 4 3 R3 = × 122 × 6.75 or, R3 = 144 × 6.75 × 3 4 or, R3 = 729 R = 3 729 = 9 cm The radius of the ball = 9 cm. (iii) Here, radius of cylindrical jar (r) = 6 cm level of oil to raise (h) = 2 cm Now, volume of oil to raise (V1) = r 2 h = × 62 × 2 = × 72 Again, radius of iron sphere (r) = 1.5 cm volume of one iron sphere (V2) = 4 3 r 3 = 4 3 × (1.5)3 = 4 3 × 3.375 No. of iron sphere (n) = V1 V2 = × 72 4 3 × 3.375 = 16.
169 Additional Practice – III 1. The area of the four walls of a cuboidal room of a school is 81 square meters. The length and breadth of the room are 8 meters and 5 meters respectively. The room has one door with area 3 square meters and 2 windows with total area 2.25 square meters. (a) Write the formula to find the area of four walls. (b) Find the height of the room. (c) What is the total cost of painting the four walls of room excluding windows and door at the rate of Rs. 175 per square meter? Find it. (d) If the paper is pasted at the rate of Rs. 250 per square meter excluding the windows and door, by how much more rupees does it require than it is colored, compare it. Solution: (a) Area of 4 walls = 2h( + b) (b) Given, length of room () = 8 m area of 4 walls = 81 m2 breadth of room (b) = 5 m Next, area of 4 walls (A) = 81 – (1 × 3 + 2 × 2.25) = 81 – (3 + 4.5) = 81 – 7.5 = 73.5 m2 Now, A = 2h( + b) or, 73.5 m2 = 2h(8 + 5) or, h = 73.5 2 × 13 = 2.83 m height of the room (h) = 2.83 m (c) Cost of painting = A × R = 73.5 × 175 = Rs. 12,862.50. (d) Cost of paper pasting = A × R = 73.5 × 250 = Rs. 18,375 Required more money = 18,375 – 12,862.5 = Rs. 5,512.50 2. A house has a room of 8 m long, 6 m wide and 4m high, with one door of 2.5 m × 1 m and two windows of 1.5 m × 2 m. (a) Write the formula for finding the area of celing. (b) How much is the area of the four walls of the room. Calculate it. (c) Compare the area of the celing and the area of the windows and the door? (d) Find the total cost of painting in the four walls of room excluding windows and door at the rate of Rs. 45 per square meter? Find it. Solution: (a) Area of ceiling = × b (b) Area of 4 walls (A) = 2h( + b) – Area of doors and windows = 2 × 4(8 + 6) – (2.5 m × 1 m + 1.5 × 2 m) = 8 × 14 – (2.5 + 3) = 112 – 5.5 = 106.5 m2
170 (c) Area of ceiling = × b = 8 × 6 = 48 m2 Area of door and window = 2.5 m × 1 m + 1.5 m × 2 m = 2.5 + 3 = 5.5 m2 Now, 48 m2 = 5.5 m2 = 42.5 m2 Area of ceiling is 42.5 m2 more. (d) Cost for painting = A × R = 106.5 × 45 = Rs. 4,792.50. 3. The curved surface area of the cylinder is 748 sq. cm and the sum of radius and height is 24 cm. (a) Write the formula to find the curved surface area of cylinder. (b) Find the volume of cylinder. (c) How much total surface area is greater than curved surface area of the cylinder? Solution: Here, CSA = 748 cm2 r + h = 24 cm (a) CSA = 2rh (b) CSA = 748 or, 2rh = 748 or, 2 × 22 7 × r (24 – r) = 748 or, 24r – r2 = 17 × 7 or, r2 – 24r + 119 = 0 or, r2 – 17r – 7r + 119 = 0 or, r(r – 17 – 7(r – 17) = 0 or, (r – 17) (r – 7) = 0 Either, r – 17 = 0 r = 17 cm OR, r – 7 = 0 r = 7 cm When, r = 17, h = 7 cm When, r = 7, h = 17 cm When, r = 17 cm and h = 7 cm Volume of cylinder (V) = r 2 h = 22 7 × 172 × 7 = 6358 cm3 . When, r = 7 and h = 17 cm V = r 2 h = 22 7 × 72 × 17 = 2618 cm3 (c) TSA = 2r(r + h) = 2 × 22 7 × 7(7 + 17) = 44 × 24 = 1056 cm2 Again, TSA = 2r(r + h) = 2 × 22 7 × 17(17 + 7) = 44 × 17 × 24 7 = 2564.57 cm2 Difference, TSA – CSA = 1056 – 748 = 308 cm2 or TSA – CSA = 2564.57 – 748 = 1816.57 cm2
171 4. A prism is shown below in the figure. (a) Write the formula for finding the total surface area of the prism. (b) Find the volume of the prism. (c) Which is greater in the lateral surface area of the prism and the area of the cross section? Find it. Solution: (a) TSA = LSA + 2 × Area of cross section area (b) Volume = CSA × H = 1 2 × 6 × 4 × 15 = 12 × 15 = 180 cm3 (c) CSA = 1 2 × b × h = 1 2 × 6 × 4 = 12 cm2 LSA = P × H = (5 + 5 + 6) × 15 = 240 cm2 Lateral surface area is greater by 240 – 12 = 228 cm2 . 5. In the given two cylinder with their measurement, the sum of diameters and heights are equal. (a) Write the formula to find the base area of cylinder. (b) Are the total surface areas of both cylinders equal? Compare it. (c) Are the volumes of both cylinder equal? Compare it. Solution: (a) Base area of cylinder = r 2 (b) TSA of 1st = 2r(r + h) = 2 × 22 7 × 3.5(3.5 + 14) = 385 cm2 TSA of 2nd = 2r(r + h) = 2 × 22 7 × 7 (7 + 7) = 44 × 14 = 616 cm2 No, they are not equal. (c) Volume of 1st = r 2 h = 22 7 × (3.5)2 × 14 = 539 cm3 Volume of 2st = r 2 h = 22 7 × 72 × 7 = 1078 cm3 No, their volumes are not equal. 6. The length of the edge of the isosceles triangular surface is 10 feet and two equal side is 11 feet with the length of the tent is 12 feet. (a) Find the total surface area of triangular faces of the tent. (b) What should be the capacity of the tent? (c) The cost of cloths of the wide 3.75 ft of the tent is Rs. 15 per feet. Is Rs. 1500 sufficient for the cloths? Calculate and write with reason. Solution: (a) TSA = LSA + 2 × CSA = (11 + 11 + 10) × 12 + 2 × 49 = 32 × 12 + 98 = 482 ft2 (b) CSA = b 4 4a2 – b2 = 10 4 4 × 112 – 102 = 5 2 484 – 100 = 5 2 × 19.6 = 49 ft2 Capacity = CSA × H = 49 × 12 = 588 ft3 . (c) Length of cloth = CSA/b = 49/3.75 = 13.07 ft and its cost = Rs. 15 × 13.07 = Rs. 196. Yes, Rs. 1500 is sufficient for the cloths.
172 7. In the figure a cylindrical can is shown. The radius and height of that can are 14 cm and 21 cm respectively. (a) Write the formula to find the total surface area of a cylinder. (b) What is the cost of coloring on the surface area of that cylindrical can at the rate of 50 paisa per square cm? Find it. (c) Is there a difference in the full capacity of the cylinder if the top of the cylinder is covered or not? Write with reasons. Solution: (a) TSA of cylinder = CSA + 2 × Area of base = 2rh + 2r 2 = 2r(r + h) (b) Given, radius (r) = 14 cm height (h) = 21 cm CSA = 2r(r + h) = 2 × 22 7 × 14 (14 + 21) = 3080 cm2 Cost of coloring = CSA × R = 3080 × 50 paisa = 154000 paisa = Rs. 1540 (c) No, there is not any difference in capacity because it is just lids not any contains space. 8. The volume of a triangular prism is 240 cm3 . The lengths of its triangular bases are 6 cm, 8 cm and 10 cm respectively. (a) Find the length of the triangular prism. (b) Find the ratio of the total surface area of the triangular prism to its rectangular surface area. Calculate. (c) Is the base of the triangle being right angled triangle? Justify with reason. Solution: (a) Here, volume of prism (V) = 240 cm3 Length of sides of triangle = 6 cm, 8 cm, 10 cm So, it forms right angled triangle CSA = 1 2 × b × h = 1 2 × 6 × 8 = 24 cm2 . Now, V = CSA × L or, 240 = 24 × L or, L = 10 cm. (b) LSA = P × L = (6 + 8 + 10) × 10 = 34 × 10 = 340 cm2 TSA = LSA + 2 × CSA = 340 + 2 × 24 = 340 + 48 = 388 cm2 Now, TSA : LSA = 388 : 340 = 199 : 170. (c) Yes, base is right angled triangle, 102 = 62 + 82 or, 100 = 100 9. The diameter of the circular part of the hemispherical bowl is 10.5 cm. (a) Write the formula for finding total surface area of the hemisphere. (b) Find the differences between the total surface area and curved surface area of the hemispherical bowl. (c) Is the capacity of bowl 300 cm3 ? Calculate. Solution: (a) TSA of hemisphere = 3r 2 (b) Diameter (d) = 10.5 cm
173 r = d 2 = 10.5 2 = 5.25 cm Now, CSA = 2r 2 = 2 × 22 7 × (5.25)2 = 173.25 cm2 TSA = 3r 2 = 3 × 22 7 × (5.25)2 = 259.88 cm2 Difference = TSA – CSA = 259.88 – 173.25 = 86.63 cm2 (c) Capacity of hemisphere (V) = 2 3 r 3 = 2 3 × 22 7 × (5.25)3 = 303.19 cm3 No, capacity is not 300 cm3 . 10. In the below two advertisement boards A and B, the length of sides of triangles and cost of painting in the board are given in the figure. (a) Write the formula for finding area of scalene triangle. (b) Find the area covered by advertisement boards A and B. (c) Compare the cost of painting in the advertisement boards A and B. Solution: (a) A = s(s – a) (s – b) (s – c) where, s = a + b + c 2 (b) Board A s = a + b + c 2 = 20 + 24 + 30 2 = 37 m Now, Area = s(s – a) (s – b) (s – c) = 37(37 – 20) (37 – 24) (37 – 30) = 37 × 17 × 13 × 7 = 57239 = 239.25 m2 Board B, Area = 1 2 bh = 1 2 × 24 × 18 = 216 m2 (c) For A, Cost = A × R = 239.25 × 250 = Rs. 59,812.50 For B, Cost = A × R = 216 × 250 = Rs. 54,000 59,812.50 – 54000 = 5,812.50 more on A.
174 UNIT IV ALGEBRA CHAPTER SEQUENCE AND SERIES 8 8.1 Introduction to Sequence and Its General Terms PRACTICE 8.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Write the definition of sequence. (b) Write the formula to compute the nth term of an arithmetic sequence. (c) What is the standard form of quadratic sequence? Solution: (a) Sequence is list of numbers which are in general term. (b) nth term of an arithmetic sequence is, tn = a + (n – 1)d where, a = first term, d = common difference (c) Standard form of quadratic sequence is tn = an2 + bn + c 2. Circle the correct answer. (a) A sequence is a finite sequence if it has ............ terms. (i) infinite (ii) finite (iii) uncountable (iv) complete (b) Which is the general form of a quadratic sequence ? (i) an + b (ii) an + b (iii) 2an + bn + c (iv) an2 + bn + c (c) What is the next term of the sequence 2, 5, 8, 11, 14, ......... (i) 3 (ii) 17 (iii) 20 (iv) 16 (d) What is the next term of the sequence 1, 3, 6, 10, 15, ......... (i) 18 (ii) 19 (iii) 20 (iv) 21 (e) What is the 3rd term of the sequence having nth term tn = 3n – 4? (i) 9 (ii) - 1 (iii) 5 (iv) 4 Solution: (a) (ii) (b) (iv) (c) (ii) (d) (iv) (e) (iii)
175 Check Your Performance 3. Observe the following general term of the sequence and answer the following questions. (a) tn = 2n + 5 (b) tn = 2 – 7n (c) an = (n + 1) (2n – 1) (d) un = (3n – 1)2 (e) tn = 2n2 + 4 (f) an = 1 – 2n 3n + 1 (g) un = (–1)n + 1 (2n2 – 1) (h) tn = (–1)n (3n + 1) 2n (i) un = (–1)n (n2 + 1) (i) What is a sequence? Define it. (ii) Write down the first five terms of the sequences. (iii) Find the difference of the term from each successive term. Solution: (a) Here, tn = 2n + 5 (i) It is arithmetic sequence (ii) n 1 2 3 4 5 tn = 2n + 5 7 9 11 13 15 So, 7, 9, 11, 13, 15, ........... (iii) Common difference = 9 – 7 = 11 – 9 = 2 (b) Here, tn = 2 – 7n (i) It is arithmetic sequence (ii) n 1 2 3 4 5 tn = 2 – 7n – 5 – 12 – 19 – 26 – 33 So, – 5, – 12, – 19, – 29, – 33, .... (iii) Common difference = – 12 + 5 = – 19 + 12 = – 7 (c) Here, an = (n + 1) (2n – 1) (i) It is quadratic sequence (ii) n 1 2 3 4 5 an = (n + 1) (2n – 1) 2 9 20 35 54 So, 2, 9, 20, 35, 54, ... (iii) As it is not arithmetic sequence, there is no common difference. (d) Here, un = (3n – 1)2 (i) It is quadratic sequence. (ii) n 1 2 3 4 5 un = (3n – 1)2 4 25 64 121 196 So, 4, 25, 64, 121, 196, ... (iii) Since it is not arithmetic sequence, there is no common difference. (e) Here, tn = 2n2 + 4 (i) It is quadratic sequence. (ii) n 1 2 3 4 5 tn = 2n2 + 4 6 12 22 36 54 So, 6, 12, 22, 36, 54, ... (f) Here, an = 1 – 2n 3n + 1 Then, n 1 2 3 4 5 an = 1 – 2n 3n + 1 – 1 4 – 3 7 – 5 10 – 7 13 – 9 16
176 So, – 1 4 , – 3 7 , – 5 10 , – 7 13 , – 9 16 , ... (g) Here, un = (–1)n + 1 (2n2 – 1) Then, n 1 2 3 4 5 un = (–1)n + 1 (2n2 – 1) 1 – 7 17 – 31 49 So, 1, – 7, 17, –31, 49, ... (h) Here, tn = (–1)n (3n + 1) 2n Then, n 1 2 3 4 5 tn = (–1)n (3n + 1) 2n – 2 7 4 – 5 3 13 8 – 8 5 (i) Here, un = (–1)n (n2 + 1) Then, n 1 2 3 4 5 un = (–1)n (n2 + 1) – 2 5 – 10 17 – 26 So, – 2, 5, – 10, 17, – 26, ... 4. Study the patterns of the terms of each sequence and answer the following questions. (a) 8, 11, 14, 17 .... (b) 27, 29, 31, 33, .... (c) 50, 45, 40, 35, .... (i) What is the general term of a sequence? Define it. (ii) Find the general term of the given sequences. (iii) Find the 15th term of each sequence. Solution: (a) 8, 11, 14, 17, ... (i) 8 = 3 × 1 + 5 11 = 3 × 2 + 5 14 = 3 × 3 + 5 17 = 3 × 4 + 5 (ii) General term tn = 3n + 5 (iii) t15 = 3 × 15 + 5 = 50 (b) 27, 29, 31, 33, ... (i) 27 = 2 × 1 + 25 29 = 2 × 2 + 25 31 = 2 × 3 + 25 33 = 2 × 4 + 25 (ii) tn = 2n + 25 (iii) t15 = 2 × 15 + 25 = 55 (c) 50, 45, 40, 35, ... (i) 50 = – 5 × 1 + 55 45 = – 5 × 2 + 55 40 = – 5 × 3 + 55 35 = – 5 × 4 + 55 (ii) tn = – 5n + 55 (iii) t15 = – 5 × 15 + 55 = – 20
177 i.e. 1 7 17 31 6 10 14 4 4 5. Observe the following sequences and write the answers of the questions. (a) 4, 7, 12, 19, .... (b) 1, 7, 17, 31, .... (c) 2, 1, –2, –7, ... (i) What is an arithmetic sequence? (ii) Add next three terms in each sequence. (iii) Find the general term of the given sequences. Solution: (a) 4, 7, 12, 19, ... (i) A sequence with constant difference between any one term and its preceding term is called an Arithmetic Sequence (ii) 4, 7, 12, 19, 28, 39, 52. The next three terms are 28, 39, 52. (iii) The general quadratic sequence is, tn = an2 + bn + c we know, t1 = a + b + c = 4 ............... (1) t2 = 4a + 2b + c = 7 ........... (2) and t3 = 9a + 3b + c = 12 ......... (3) solving (1) and (2) 3a + b = 3 ........................... (4) solving (2) and (3) 5a + b = 5 ........................... (v) solving (4) and (5) 2a = 2 a = 1 and b = 0 From equation (1) 1 + 0 + c = 4 c = 3 Now, tn = a2 + 3 (b) 1, 7, 17, 31, ... (i) In this sequence, there is common difference in sequence of difference of given sequence. So it is quadratic sequence. (ii) 1, 7, 17, 31, 49, 71, 97 (iii) The general quadratic sequence is tn = an2 + bn + c we know, t1 = a + b + c = 1 ............ (1) t2 = 4a + 2b + c = 7 ........ (2) and t3 = 9a + 3b + c = 17 ...... (3) Solving (1) and (2) 3a + b = 6 ...................... (4) Solving (2) and (3) 5a + b = 10 ................... (5) Solving (4) and (5) 2a = 4 a = 2 and b = 0 From equation (1) 2 + 0 + c = 1 c = – 1 Hence, tn = 2n2 – 1
178 (c) 2, 1, – 2, – 7, ... (i) In this sequence, there is common difference in sequence of difference of given sequence. So it is quadratic sequence. (ii) 2, 1, – 2, – 7, – 14, – 23, – 34 (iii) The general quadratic sequence is tn = an2 + bn + c we have t1 = a + b + c = 2 ......... (1) t2 = 4a + 2b + c = 1 ..... (2) and t3 = 9a + 3b + c = – 2 ... (3) Solving (1) and (2) 3a + b = – 1 ................... (4) Solving (2) and (3) 5a + b = – 3 .................... (5) Solving (4) and (5) 2a = – 2 a = – 1 and b = 2 From equation (1) – 1 + 2 + c = 2 c = 1 Hence, tn = – n2 + 2n + 1 6. Study the following sequences and write the answers of the questions. (a) 1, 2, 3, 4, .... (b) 8, 3, – 2, –7, .... (c) 5, 11, 19, 29, ... (d) 3 5 , 4 7 , 6 9 , 9 11 , ... (e) 0.28, 0.35, 0.42, 0.49, ... (f) 3 2048 , 3 1024 , 3 512 , 3 256 , ... (i) Write any one difference between finite and infinite sequences. (ii) Find the nth term of the given sequences. (iii) Evaluate the 30th term and 45th term. Solution: (a) Here, 1, 2, 3, 4, ... (i) In finite sequence, there is last term and in infinite sequence, there is no last term. (ii) tn = a + (n – 1)d Here, a = 1, d = 2 – 1 = 3 – 2 = 1 So, tn = 1 + (n – 1) 1 = 1 + n – 1 = n (iii) t30 = 30 t45 = 45 (b) (ii) Here, 8, 3, –2, –7, ... a = 8 d = 3 – 8 = – 2 – 3 = – 5 So, tn = a + (n – 1)d = 8 + (n – 1) (– 5) = 8 – 5n + 5 = 13 – 5n (iii) t30 = 13 – 5 × 30 = – 137 t40 = 13 – 5 × 45 = – 212 (c) Here, 5, 11, 19, 29, ... (ii) As there is second common difference, it is quadratic sequence. The general equation is tn = an2 + bn + c
179 we know, t1 = a + b + c = 5 ........... (1) t2 = 4a + 2b + c = 11 ..... (2) t3 = 9a + 3b + c = 19 ..... (3) Solving (1) and (2) 3a + b = 6 ...................... (4) Solving (2) and (3) 5a + b = 8 ...................... (5) Solving (4) and (5) 2a = 2 a = 1 and 3 × 1 + b = 6 b = 3 From equation (i) 1 + 3 + c = 5 c = 1 Hence, tn = n2 + 3n + 1 (iii) t30 = 302 + 3 × 30 + 1 = 991 t45 = 452 + 3 × 45 + 1 = 2161 (d) (ii) Here, 3 5 , 4 7 , 6 9 , 9 11 , ... In numerator, due to the second common difference, it is quadratic sequence. And the general term (tn) = an2 + bn + c we know, t1 = a + b + c = 3 ...... (1) t2 = 4a + 2b + c = 4 ..... (2) t3 = 9a + 3b + c = 6 ..... (3) Solving (1) and (2) 3a + b = 1 .................... (4) Solving (2) and (3) 5a + b = 2 .................... (5) Solving (4) and (5) 2a = 1 a = 1 2 and 3 × 1 2 + b = 1 b = – 1 2 From (1) 1 2 – 1 2 + c = 3 c = 3 So, tn = 1 2 n2 – 1 2 n + 3 In case of denominator, 5, 7, 9, 11, ... a = 5 d = 7 – 5 = 9 – 7 = 2 tn = a + (n – 1)d = 5 + (n – 1)2 = 5 + 2n – 2 = 2n + 3
180 Hence, tn = 1 2 n2 – 1 2 n + 3 2n + 3 (iii) t30 = 1 2 × 302 – 1 2 × 30 + 3 2 × 30 + 3 = 438 63 (e) Here, 0.28, 0.35, 0.42, 0.49, ... (ii) Here, a = 0.28 d = 0.35 – 0.28 = 0.42 – 0.35 = 0.07 tn = a + (n – 1)d = 0.28 + (n – 1) 0.07 = 0.28 + 0.07n – 0.07 = 0.21 + 0.07n (iii) t30 = 0.21 + 0.07 × 30 = 0.21 + 2.1 = 2.31 t45 = 0.21 + 0.07 × 45 = 3.36 (f) 3 2048 , 3 1024 , 3 512 , 3 256 , ... (ii) Here, numerator is constant 3, and denominator is 2048, 1024, 512, 256, ... a = 2048 r = 1024 2048 = 512 1024 = 1 2 tn = 3 × arn – 1 = 3 × 2048 1 2 n – 1 = 3 × 2048 1 2n – 1 × 3 × 211 × 21 – n = 3 × 2048 × 2 2n 3 × 211 – n (iii) t30 = 3 × 212 – 30 = 3 × 2–18 = 3 262144 t45 = 3 × 212 – 45 = 3 × 2–33 = 3 8589934592 7. Observe the following patterns of the sequences and answer the questions. (a) (i) What is a quadratic sequence? Define it. (ii) Add the next one figure on each sequence. (iii) Find the nth term of the given sequence. (iv) Find the number of triangles in the 25th figure. (b) (i) Draw the next one shape on the sequence. (ii) Find the nth term of the given sequence. (iii) Find the number of glasses in the 20th shape. (c) (i) Construct next one shape on the sequence. (ii) Find the nth term of the given sequences. (iii) Find the number of vertices in the 30th shape. Solution: (a) (i) Quadratic sequence is a sequence in which the second difference of any term and its preceding term is constant. (ii) The required next one figure is, (iii) Sequence of no. of triangles is 1, 4, 7, 10, 13, ... Here, a = 1 d = 4 – 1 = 7 – 4 = 3 tn = a + (n – 1)d = 1 + (n – 1) × 3 = 1 + 3n – 3 = 3n – 2
181 (iv) when n = 25 t25 = 3 × 25 – 2 = 75 – 2 = 72 (b) (i) The required next one figure is, (ii) Sequence of cone is 1, 3, 6, 10, 15, ... The sequence has arithmetic sequence of difference of given sequence. So it is quadratic sequence. Now, tn = an2 + bn + c we know, t1 = a + b + c = 1 ........ (1) t2 = 4a + 2b + c = 3 .... (2) and t3 = 9a + 3b + c = 6 .... (3) Solving (1) and (2) 3a + b = 2 .................... (4) Solving (2) and (3) 5a + b = 3 ..................... (5) Solving (4) and (5) 2a = 1 a = 1 2 and b = 2 – 3a = 2 – 3 × 1 2 = 1 2 From equation (1) 1 2 + 1 2 + c = 1 c = 0 Hence, tn = 1 2 n2 + 1 2 n + 0 = 1 2 n(n + 1) (iii) When, n = 20 t20 = 1 2 × 20(20 + 1) = 210 (c) (i) The required next one figure is, (ii) Sequence of dots is 1, 2, 4, 8, 16, ... It is in geometric sequence with a = 1 r = 2 1 = 4 2 = 2 and, tn = arn – 1 or, tn = 1 × 2n – 1 = 2n – 1 (iii) when, n = 30 t30 = 230 – 1 = 229 = 536870912
182 8.2 Introduction to Series PRACTICE 8.2 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define series. (b) What is sigma notation? Solution: (a) Series is the sequence representing in form of sum. For example, if t1, t2, t3, t4, ..., tn then, series, t1 + t2 + t3 + t4 + ... + tn (b) If the sequence is notify is sum of all terms together, it is called sigma notation. For example: t1, t2, t3, t4, ..., tn Sn = n = 1 n tn 2. Represent the following sequences into their associated series: (a) 1, 6, 11, 16, .... to n terms (b) 15, 6, –3, –12, .... to n terms (c) 15, 20, 30, 45, .... to n terms (d) 45, 39, 31, 21, .... to n terms (e) 2, 4, 6, 8, .... to n terms (f) 5, 10, 20, 40, .... to n terms Solution: (a) Here, 1, 6, 11, 16, ... to n terms then, 1 + 6 + 11 + 16 + ... + nth term (b) Here, 15, 6, – 3, – 12, ... to n terms then, 15 + 6 – 3 – 12, ... – nth term (c) Here, 15, 20, 30, 45, ... to n terms then, 15 + 20 + 30 + 45 + ... + nth terms (d) Here, 45, 39, 31, 21, ... to n terms then, 45 + 39 + 31 + 21 + ... + nth terms (e) Here, 2, 4, 6, 8, ... to n terms then, 2 + 4 + 6 + 8 + ... + nth terms (f) Here, 5, 10, 20, 40, ... to n terms then, 5 + 10 + 20 + 40 + ... + nth terms Check Your Performance 3. Observe the nth term of some sequences below where n ∈ N. (a) tn = 2n2 – n + 3 (b) an = 2n + 3 2n – 1 (c) un = (–1)n – 1 3n (d) tn = (–1)n 3n (i) What is nth term ? Define it. (ii) Write the first five terms of each sequence. (iii) Find the sum of the first five terms (S5) of each sequence. Solution: (a) Here, tn = 2n2 – n + 3 (i) nth term is the term of required position on given equation of general term.
183 (ii) n 1 2 3 4 5 tn = 2n2 – n + 3 4 9 18 31 48 (iii) S5 = 4 + 9 + 18 + 31 + 48 = 110 (b) Here, an = 2n + 3 2n – 1 (ii) n 1 2 3 4 5 an = 2n + 3 2n – 1 5 7 3 9 5 11 7 13 9 (iii) S5 = 5 + 7 3 + 9 5 + 11 7 + 13 9 = 3827 315 = 12 47 315 (c) Here, un = (–1)n – 1 3n (ii) n 1 2 3 4 5 un = (–1)n – 1 3n 3 – 9 27 – 81 243 (iii) S5 = 3 – 9 + 27 – 81 + 243 = 183 (d) Here, tn = (–1)n 3n (ii) n 1 2 3 4 5 tn = (–1)n 3n – 3 9 – 27 81 – 243 (iii) S5 = –3 + 9 – 27 + 81 – 243 = – 183 4. Observe the summation or sigma notation and answer the following questions. (a) n = 1 4 (2n2 – 2n + 3) (b) n = 2 6 (2n + 1) (3n – 1) (c) n = 4 7 2n2 + 3 n – 2 (d) n = 2 6 (–1)n (2n + 1) (e) n = 3 8 (–1)n+1 (n2 + 2) (f) n = 2 7 (– 1)n (2n + 1) 3n – 1 (i) What are the values of n in the given sigma notations? (ii) Write the expanded form of the given summation form. (iii) Find their values. Solution: (a) n = 1 4 (2n2 – 2n + 3) = (2 × 12 – 2 × 1 + 3) + (2 × 22 – 2 × 2 + 3) + (2 × 32 – 2 × 3 + 3) + (2 × 42 – 2 × 4 + 3) = 3 + 7 + 15 + 27 = 52 (b) n = 2 6 (2n + 1) (3n – 1) = (2 × 2 + 1) (3 × 2 – 1) + (2 × 3 + 1) (3 × 3 – 1) + (2 × 4 + 1) (3 × 4 – 1) + (2 × 5 + 1) (3 × 5 – 1) + (2 × 6 + 1) (3 × 6 – 1) = 25 + 56 + 99 + 154 + 221 = 555 (c) n = 4 7 2n2 + 3 n – 2 = 2 × 42 + 3 4 – 2 + 2 × 52 + 3 5 – 2 + 2 × 62 + 3 6 – 2 + 2 × 72 + 3 7 – 2 = 35 2 + 53 3 + 75 4 + 101 5 = 4447 60
184 (d) n = 2 6 (–1)n (2n + 1) = (–1)2 (2 × 2 + 1) + (–1)3 (2 × 3 + 1) + (–1)4 (2 × 4 + 1) + (–1)5 (2 × 5 + 1) + (–1)6 (2 × 6 + 1) = 5 – 7 + 9 – 11 + 13 = 9 (e) n = 3 8 (–1)n+1 (n2 + 2) = (–1)3 + 1 (32 + 2) + (–1)4 + 1 (42 + 2) + (–1)5 + 1 (52 + 2) + (–1)6 + 1 (62 + 2) + (–1)7 + 1 (72 + 2) + (–1)8 + 1 (82 + 2) = 11 – 18 + 27 – 38 + 51 – 66 = – 33 (f) n = 2 7 (– 1)n (2n + 1) 3n – 1 = (– 1)2 (2 × 2 + 1) 3 × 2 – 1 + (– 1)3 (2 × 3 + 1) 3 × 3 – 1 + (– 1)4 (2 × 4 + 1) 3 × 4 – 1 + (– 1)5 (2 × 5 + 1) 3 × 5 – 1 + (– 1)6 (2 × 6 + 1) 3 × 6 – 1 + (– 1)7 (2 × 7 + 1) 3 × 7 – 1 = 5 5 – 7 8 + 9 11 – 11 14 + 13 17 – 15 20 = 1803 10472 5. Observe the following sigma notation and answer the following questions. (a) n = 2 16 xn (b) n = 3 25 (2n2 + 3) (c) k = 5 20 (–1)k 4k – 1 k + 1 (i) How many terms are there in the series as in the given sigma notation. (ii) Find its 12th term. Solution: (a) Here, n = 2 16 xn (i) No. of terms (n) = 16 – 2 + 1 = 15 (ii) t12 = x12 + 1 [ Here 2 – 1 = 1 adding on term] = x13 (b) Here, n = 3 25 (2n2 + 3) (i) No. of terms (n) = 25 – 3 + 1 = 23 (ii) t12 = 2(12 + 2)2 + 3 [ Here 3 – 1 = 2 adding on term] = 2 × 196 + 3 = 395 (c) Here, k = 5 20 (–1)k 4k – 1 k + 1 (i) No. of terms (n) = 20 – 5 + 1 = 16 (ii) t12 = (–1)12 + 4 4(12 + 4) – 1 (12 + 4) + 1 [ Here 5 – 1 = 4 adding on term] = (–1)16 64 – 1 17 = 63 17 6. Write each of the following by using sigma notation: (a) 4 + 7 + 10 + 13 + 16 + 19 + 22 (b) 2 5 + 6 7 + 10 9 + 14 11 + 18 13 + 22 15 (c) – 3 5 + 5 9 – 7 13 + 9 17 – 11 21 (d) 1 + 5 + 25 + 125 + 625 + 3125
185 (e) (r – 1) 2 + (r – 1)2 4 + (r – 1)3 8 + (r – 1)4 16 + (r – 1)5 32 + (r – 1)6 64 (f) a + ar + ar2 + ar3 + ar4 + ar5 + ar6 (i) Find the nth term of the given series. (ii) Write each sequence by using sigma notation. Solution: (a) Here, 4 + 7 + 10 + 13 + 16 + 19 + 22 a = 4 and d = 7 – 4 = 3 So, General term (tn) = a + (n – 1)d = 4 + (n – 1) × 3 = 4 + 3n – 3 = 3n + 1 and Sn = n = 1 7 3n + 1 (b) Here, 2 5 + 6 7 + 10 9 + 14 11 + 18 13 + 22 15 In numerator, a = 2 and d = 6 – 2 = 4 tn = a + (n – 1)d = 2 + (n – 1) × 4 = 2 + 4n – 4 = 4n – 2 In denominator, a = 5 and d = 7 – 5 = 2 tn = a + (n – 1)d = 5 + (n – 1) × 2 = 5 + 2n – 2 = 2n + 3 Here, total number of term (n) = 6 So, Sn = n = 1 6 4n – 2 2n + 3 (c) Here, – 3 5 + 5 9 – 7 13 + 9 17 – 11 21 The terms are is alternate sign. As, 1st term is +ve, (–1)n is co-efficient of each term. then, in numerator, a = 3 and d = 5 – 3 = 2 tn = a + (n – 1)d = 3 + (n – 1) × 2 = 3 + 2n – 2 = 2n + 1 In denominator, a = 5 and d = 9 – 5 = 4 tn = a + (n – 1)d = 5 + (n – 1) × 4 = 5 + 4n – 4 = 4n + 1 Total number of term (n) = 5 So, Sn = n = 1 5 (–1)n 2n + 1 4n + 1 (d) Here, 1 + 5 + 25 + 125 + 625 + 3125 a = 1 and r = 5 1 = 25 5 = 5 then, general term (tn) = arn – 1 = 1 (5)n – 1 = 5n – 1 And total no. of terms (n) = 6 So, Sn = n = 1 6 5n – 1 (e) Here, (r – 1) 2 + (r – 1)2 4 + (r – 1)3 8 + (r – 1)4 16 + (r – 1)5 32 + (r – 1)6 64 In numerator, a = r – 1 common ratio (r) = (r – 1)2 r – 1 = r – 1 (i) (ii)
186 then, general term (tn) = arn – 1 = (r – 1) (r – 1)n – 1 = (r – 1)n Again, in denominator, a = 2 common ratio (r) = 4 2 = 8 4 = 2 then, general term (tn) = arn – 1 = 2 × 2n – 1 = 2n And, total number of terms (n) = 6 Hence, Sn = n = 1 6 (r – 1)n 2n (f) Here, a + ar + ar2 + ar3 + ar4 + ar5 + ar6 a = a common ratio (r) = ar a = ar2 ar = r general term (tn) = arn – 1 = arn – 1 And, Sn = n = 1 6 arn – 1 7. Observe the following series whose sum of the first n terms are given below: (a) Sn = 2n + 1 (b) Sn = 3n2 – 1 (c) Sn = (2n – 1)2 (d) Sn = 2n2 + 4n (i) Find the sum of the first four terms from the given sum of n terms. (ii) Find the sum of the first 25 terms. (iii) Find the sum of the first (n – 1) terms from the given sum of the first n terms. (iv) Find the nth term from the given sum of n terms. Solution: (a) Here, Sn = 2n + 1 (i) S4 = 2 × 4 + 1 = 9 (ii) S25 = 2 × 25 + 1 = 51 (iii) Sn – 1 = 2(n – 1) + 1 = 2n – 2 + 1 = 2n – 1 (iv) tn = Sn – Sn – 1 = 2n + 1 – (2n – 1) = 2n + 1 – 2n + 1 = 2 (b) Here, Sn = 3n2 – 1 (i) S4 = 3 × 42 – 1 = 47 (ii) S24 = 3 × 252 – 1 = 1874 (iii) Sn – 1 = 3(n – 1)2 – 1 = 3n2 – 6n + 3 – 1 = 3n2 – 6n + 2 (iv) tn = Sn – Sn – 1 = 3n2 – 1 – (3n2 – 6n + 2) = 3n2 – 1 – 3n2 + 6n – 2 = 6n – 3 (c) Here, Sn = (2n – 1)2 (i) S4 = (2 × 4 – 1)2 = 72 = 49 (ii) S25 = (2 × 25 – 1)2 = 492 = 2401 (iii) Sn – 1 = {2(n – 1) – 1}2 = (2n – 2 – 1)2 = (2n – 3)2 (iv) tn = Sn – Sn – 1 = (2n – 1)2 – (2n – 3)2 = (2n – 1 – 2n + 3) (2n – 1 + 2n – 3) = 2(4n – 4) = 8n – 8 (d) Here, Sn = 2n2 + 4n (i) S4 = 2 × 42 + 4 × 4 = 48 (ii) S25 = 2 × 252 + 4 × 25 = 1350 (iii) Sn – 1 = 2(n – 1)2 + 4(n – 1) = 2n2 – 4n + 2 + 4n – 4 = 2n2 – 2 (iv) tn = Sn – Sn – 1 = 2n2 + 4n – 2n2 + 2 = 4n + 2
187 8.3 General Term of Arithmetic Sequence PRACTICE 8.3 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define arithmetic sequence. (b) In which condition does a sequence become arithmetic? (c) What is the common difference in the sequence a + d, a, a – d, a – 2d, .............? Solution: (a) The sequence having common difference is called arithmetic sequence. (b) The sequence between arithmetic if it get common difference. (c) In sequence, a + d, a, a – d, a – 2d, ... has common difference = a – (a + d) = (a – d) – a = – d 2. (a) If the first term and the common difference of an AS are 7 and –3 respectively then find the first five terms of the AS. (b) If the first term and the common difference of arithmetic sequence are 2 3 and 1 3 respectively then find the arithmetic sequence. Solution: (a) Here, a = 7 and d = – 3 Now, t1 = a = 7 t2 = a + d = 7 – 3 = 4 t3 = a + 2d = 7 + 2 × (–3) = 1 t4 = a + 3d = 7 + 3 × (–3) = – 2 t5 = a + 4d = 7 + 4 × (–3) = – 5 (b) Here, a = 2 3 , d = 1 3 Now, t1 = a = 2 3 t2 = a + d = 2 3 + 1 3 = 1 t3 = a + 2d = 2 3 + 2 × 1 3 = 4 3 t4 = a + 3d = 2 3 + 3 × 1 3 = 5 3 t5 = a + 4d = 2 3 + 4 × 1 4 = 2 So, the sequence is, 2 3 , 1, 4 3 , 5 3 , 2, ... Check Your Performance 3. Observe the following arithmetic sequences: (i) 101, 120, 139, 158, ...... (ii) 120 + 114 + 108 + 102 + ...... (a) Write the formula to find the nth term of the arithmetic sequence. (b) Find the common difference of the arithmetic sequence. (c) Find the nth term of the arithmetic sequence. (d) Find its 20th term.
188 Solution: (i) Here, 101, 120, 139, 158, ...... (a) Formula to find the nth term (tn) = a + (n – 1)d (b) Common difference (d) = 120 – 101 = 139 – 120 = 19 (c) General term (tn) = a + (n – 1)d = 101 + (n – 1) × 19 = 101 + 19n – 19 = 19n + 82 (d) t20 = 19 × 20 + 82 = 462 (ii) Here, 120 + 114 + 108 + 102 + ...... (b) Common difference (d) = 114 – 120 = 108 – 114 = – 6 (c) General term (tn) = a + (n – 1)d = 120 + (n – 1) (– 6) = 120 – 6n + 6 = 126 – 6n (d) t20 = 126 – 6 × 20 = 6 4. (i) If the 10th term of an AP having the common difference 5 is 43, find the first term. (ii) The 12th term of an arithmetic series, whose first term is 0.35, is 0.02. Find its common difference. Solution: (i) Here, d = 5, t10 = 43 and a = ? we have, tn = a + (n – 1)d or, t10 = a + (10 –1) × 5 or, 43 = a + 45 a = – 2 (ii) Here, a = 0.35, t12 = 0.02 and d = ? we have, tn = a + (n – 1)d or, t12 = 0.35 + (12 – 1)d or, 0.02 = 0.35 + 11d or, – 0.33 = 11d d = – 0.03 5. (i) How many terms are there in the AS 2, 9, 16, 23, …………., 142 ? (ii) Which term of the arithmetic series 21 + 25 + 29 + 33 + ……. is 397? (iii) Is – 29 3 a term of the series 19 3 + 13 3 + 7 3 + ........? Solution: (i) Here, 2, 9, 16, 23, ..., 142 a = 2 d = 9 – 2 = 16 – 9 = 7 tn 142 n = ? we have, tn = a + (n – 1)d or, 142 = 2 + (n – 1) × 7 or, 142 – 2 = 7(n – 1) or, 140 2 = n – 1 or, 20 + 1 = n n = 21 (ii) Here, 21 + 25 + 29 + 33 + ... + 397 a = 21 d = 25 – 21 = 29 – 25 = 4 tn = 397 n = ?
189 we have, tn = a + (n – 1)d or, 397 = 21 + (n – 1) × 4 or, 397 – 21 = 4(n – 1) or, 376 4 = n – 1 or, 94 + 1 = n n = 95 (iii) Here, 19 3 + 13 3 + 7 3 + ... a = 19 3 d = 13 3 – 19 3 = 7 3 – 13 3 = – 2 In tn = – 29 3 , n should be whole number we have, tn = a + (n – 1)d or, – 29 3 = 19 3 + (n – 1) (– 2) or, – 29 3 – 19 3 = – 2(n – 1) or, – 16 – 2 = n – 1 or, 8 + 1 = n n = 9 N, – 29 3 is term of sequence. 6. (i) Study the first three terms 2a + 1, 3a, 5a + 1 of an AP. (ii) Study the first three terms a, a2 + 1 and a + 6 of an AS. (a) Find the value of a. (b) Find these three terms. (c) Compute the next three terms. (d) Find its nth term. Solution: (i) Here, 2a + 1, 3a and 5a + 1 are in AP (a) So, 3a – (2a + 1) = (5a + 1) – 34 or, a – 1 = 2a + 1 a = – 2 (b) And, 2a + 1 = 2(– 2) + 1 = – 3 3a = 3(– 2) = – 6 5a + 1 = 5(– 2) + 1 = – 9 (c) –3, –6, –9, –12, –15, – 18 (d) a = – 3, d = – 6 – (– 3) = – 6 + 3 = – 3 tn = a + (n – 1)d = – 3 + (n – 1) (– 3) = – 3 – 3n + 3 = – 3n (ii) Here, a, a2 + 1 and a + 6 are in AS (a) So, (a2 + 1) – a = (a + 6) – (a2 + 1) or, a2 + 1 – a = a + 6 – a2 – 1 or, 2a2 – 2a – 4 = 0 or, a2 – a – 2 = 0 or, (a – 2) (a + 1) = 0 a = – 1 or 2 (b) when, a = – 1 a = – 1 a2 + 1 = (– 1)2 + 1 = 2
190 a + 6 = – 1 + 6 = 5 when, a = 2 a = 2 a2 + 1 = 22 + 1 = 5 a + 6 = 2 + 6 = 8 (c) when, a = – 1 –1, 2, 5, 8, 11, 14 are required AS when, a = 2 2, 5, 8, 11, 14, 17 are required AS (d) when, a = – 1 tn = – 1 + (n – 1) × 3 = – 1 + 3n – 3 = 3n – 4 when, a = 2 tn = 2 + (n – 1) × 3 = 2 + 3n – 3 = 3n – 1 7. (i) The sixth and seventeenth terms of an arithmetic progression are 19 and 41 respectively. (a) Find the common difference of the AS. (b) Find its first term. (c) Is 36 a term of the sequence? (d) Find its 100th term. (ii) The 5th term and 8th term of an AP are 19 and 31 respectively. (a) Find the common difference of the AS. (b) Find its first term. (c) Which term of it is 67 ? (d) Find its 80th term. Solution: (i) Here, t6 = 19, t17 = 41 (a) we have, tn = a + (n – 1)d or, t6 = a + (6 – 1)d or, 19 = a + 5d or, 19 – 5d = a .................. (i) Also, t17 = a + (17 – 1)d or, 41 = 19 – 5d + 16d or, 41 – 19 = 11d or, 22 11 = d d = 2 (b) Putting the value of d in equation (i) a = 19 – 5 × 2 = 9 (c) If 36 is term of sequence, tn = 36 or, a + (n – 1)d = 36 or, 9 + (n – 1) × 2 = 36 or, 2(n – 1) = 27 or, n – 1 = 27 2 or, n = 13.5 + 1 = 14.5 W So, 36 is not term of sequence. (d) t100 = a + (100 – 1)d = 9 + 99 × 2 = 207 (ii) Here, t5 = 19 and t8 = 31
191 (a) we have, tn = a + (n – 1)d or, t5 = a + (5 – 1)d or, 19 = a + 4d or, a = 19 – 4d ...................... (i) Also, t8 = a + (8 – 1)d or, 31 = 19 – 4d + 7d or, 31 – 19 = 3d or, 12 3 = d d = 4 (b) Putting the value of d is equation (i) a = 19 4 × 4 = 3 (c) If tn = 67 or, a + (n – 1)d = 67 or, 3 + (n – 1)4 = 67 or, 4(n – 1) = 64 or, n – 1 = 16 n = 17 (d) t80 = a + (80 – 1)d = 3 + 79 × 4 = 319 8. (i) 8 times of the eighth term of an AP is equal to 12 times of the twelfth term. Find its first term if the common difference is – 2. (ii) 8 times of the third term and double of the tenth term of an AP are equal to each other. Find the common difference of the AP if its first term is 2. Solution: (i) Here, d = – 2 and 8t8 = 12 t12 or, 8{a + (8 – 1)d} = 12{a + (12 – 1)d} or, 8{a + 7 × (–2)} = 12{a + 11(–2)} or, 8(a – 14) = 12(a – 22) or, 2a – 28 = 3a – 66 a = 38 (ii) Here, a = 2 and 8t3 = 2t10 or, 4{a + (3 – 1)d} = a + (10 – 1)d or, 4(2 + 2d) = 2 + 9d or, 8 + 8d = 2 + 9d d = 6 9. (i) The 5th term of an AP with first term 1 40 is 1 8 . (a) Find the common difference of the AS. (b) Find its 11th term. (ii) The third term and the sixth term of an AP are 1 and 2 respectively. (a) Find the common difference of the AS. (b) Find its first term. (c) Find its 27th term. (iii) If 5 times of the fifth term is equal to 12 times of the twelfth term of an AP, find its 17th term. Solution: (i) Here, a = 1 40 , t5 = 1 8 (a) we have, tn = a + (n – 1)d
192 or, t5 = 1 40 + (5 – 1)d or, 1 8 = 1 40 + 4d or, 1 8 – 1 40 = 4d or, 4 40 = 4d d = 1 40 (b) t11 = a + (11 – 1)d = 1 40 + 10 1 40 = 11 40 (ii) Here, t3 = 1 and t6 = 2 (a) we have, tn = a + (n – 1)d or, t3 = a + (3 – 1)d or, 1 = a + 2d a = 1 – 2d ..................... (i) Also, t6 = a + (6 – 1)d or, 2 = 1 – 2d + 5d or, 1 = 3d d = 1 3 (b) putting the value of d in equation (i) a = 1 – 2 × 1 3 = 1 3 (c) t27 = a + (27 – 1)d = 1 3 + 26 × 1 3 = 27 3 = 9 (iii) Here, 5t5 = 12t12 or, 5{a + (5 – 1)d} = 12{a + (12 – 1)d} or, 5(a + 4d) = 12(a + 11d) or, 5a + 20d = 12a + 132d or, 0 = 7a + 112d or, 0 = 7(a + 16d) or, 0 = a + (17 – 1)d i.e. t17 = 0 10. (i) The sum of three successive numbers in an AP is 54 and the product of the last two terms is 333. (ii) Divide 36 into three parts which are in AP such that their product is 756. (a) Suppose the three consecutive numbers in AP. (b) Find these three numbers. Solution: (i) (a) Let a – d, a and a + d are in AP then, a – d + a + a + d = 54 or, 3a = 54 a = 18 Again, a(a + d) = 333 or, 18(18 + d) = 333 or, 18 + d = 333 18 or, d = 37 2 – 18 d = 1 2
193 Now, a – d = 18 – 1 2 = 35 2 a = 18 a + d = 18 + 1 2 = 37 2 (b) 35 2 , 18, 37 2 is AS (ii) (a) Let, a – d, a and a + d are in AP then, a – d + a + a + d = 36 or, 3a = 36 a = 12 Also, (a – d)a (a + d) = 756 or, a(a2 – d2 ) = 756 or, 12(122 – d2 ) = 756 or, 144 – d2 = 63 or, d2 = 81 d = ± 9 (b) If a = 9 a – d = 12 – 9 = 3 a = 12 a + d = 12 + 9 = 21 If a = – 9 a – d = 12 + 9 = 21 a = 12 a + d = 12 – 9 = 3 Hence, required sequence is 3, 12, 21 or 21, 12, 3. 11. (i) A man is appointed on a salary of Rs. 15500 per month. He gets an increment of Rs. 1500 as grade every year. What is his monthly salary in the fifth year? (ii) The meter of a taxi starts with a reading of Rs. 15 and then runs up Rs. 14 for each kilometer travelled. How much should be paid for a journey of 12 km? Solution: (i) Here, a = 15500 d = 1500 n = 5 Now, tn = a + (n – 1)d t5 = 15500 + (5 – 1)1500 = 15500 + 6000 = 21500 Hence, monthly salary in the fifth year = Rs. 21500 (ii) Here, a = 15 d = 14 n = 12 Now, tn = a + (n)d t12 = 15 + (12)14 = 15 + 160 = 183 Hence, total charge = 183.
194 8.4 General Term of Geometric Sequence PRACTICE 8.4 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Write the definition of geometric series. (b) What is the general term of a sequence ? (c) Write any one property of the geometric sequence. Solution: (a) Geometric series is the series with common ratio. (b) General term of GS (tn) = arn – 1 (c) The major property of the geometric sequence is having common ratio. 2. Which of the following sequences is a geometric sequence or series? (a) 1, 3, 9, 27, …….. (b) 2, 4, 6, 8, ……. Solution: (a) Here, 1, 3, 9, 27, ... Since, 3 1 = 9 3 = 27 9 = 3, it is geometric sequence. (b) Here, 2, 4, 6, 8, ... Since, 4 2 6 4 8 6 , it is not geometric sequence. Check Your Performance 3. (i) The first term and the common ratio of a geometric sequence are 32 and 1 2 respectively. (a) How can you find the common ratio of a geometric sequence? (b) Find the first six terms of the sequence. (ii) If the 6th term of a GP with the common ratio 3 is 243 2 , find its first term. Solution: (i) Here, a = 32, r = 1 2 (a) Common ratio of GS is calculated dividing second term by first term (or, successive term by precceding term) (b) t1 = a = 32 t2 = ar = 32 × 1 2 = 16 t3 = ar2 = 32 × 1 22 = 8 t4 = ar3 = 32 × 1 23 = 4 t5 = ar4 = 32 × 1 24 = 2 t6 = ar5 = 32 × 1 25 = 1 (ii) Here, r = 3 and t6 = 243 2
195 we have, tn = arn – 1 or, t6 = a × 36 – 1 or, 243 2 = a × 35 or, 243 2 = a × 243 a = 1 2 4. Observe the following geometric sequences and answer the following questions. (i) 243, 81, 27,…… (ii) 1 243 + 1 81 + 1 27 + ....... (a) Write the formula to find the general term of a geometric sequence. (b) Find the common ratio of the sequence. (c) Find the nth term of the sequence. (d) Find the 15th term of the sequence. Solution: (i) Here, 243, 81, 27, ... (a) general term of GS (tn) = arn – 1 (b) common ratio (r) = 81 243 = 27 81 = 1 3 (c) tn = 243 × 1 3 n – 1 = 35 × 1 3n – 1 = 1 3n – 6 = 1 3 n – 6 (d) t15 = 1 3 15 – 6 = 1 3 9 = 1 19683 (ii) Here, 1 243 + 1 81 + 1 27 + ... (b) r = 1 81 1 243 = 1 27 1 81 = 3 (c) tn = arn – 1 = 1 243 × 3n – 1 = 1 35 × 3n – 1 = 3n – 6 (d) t15 = 315 – 6 = 39 = 19683 5. (i) The last term of a geometric sequence is 26244. If its first term and common ratio are 4 and 3 respectively then find the number of terms. (ii) Does 0.50625 belong to the geometric series 0.1, 0.15, 0.225……….. ? Solution: (i) Here, a = 4, r = 3 and tn = 26244 Now, tn = arn – 1 or, 26244 = 4 × 3n – 1 or, 6561 = 3n 3 or, 19683 = 3n or, 39 = 3n n = 9 (ii) Here, 0.1, 0.15, 0.225, .... a = 0.1 r = 0.15 0.1 = 1.5
196 If tn = 0.50625 or, arn – 1 = 0.50625 or, 0.1 × (1.5)n – 1 = 0.50625 or, (1.5)n – 1 = 5.0625 or, (1.5)n – 1 = (1.5)4 then, n – 1 = 4 n = 5 W Hence, 0.50625 is term of sequence. 6. (i) (x + 2), (x + 10) and 9(x + 2) are three consecutive terms in GP. (ii) (x + 1), (x – 5) and 1 are in a geometric series. (a) Find the value of x. (b) Find these three terms. (c) Compute the next three terms. (d) Find its nth term. Solution: (i) (a) Here, (x + 2), (x + 10) and 9(x + 2) are in GP then, x + 10 x + 2 = 9x + 18 x + 10 or, x2 + 20x + 100 = 9x2 + 36x + 36 or, 0 = 8x2 + 16x – 64 or, x2 + 2x – 8 = 0 or, (x + 4) (x – 2) = 0 x = {–4, 2} (b) when, x = – 4 x + 2 = – 4 + 2 = – 2 x + 10 = – 4 + 10 = 6 9(x + 2) = 9(– 4 + 2) = – 18 when, x = 2 x + 2 = 2 + 2 = 4 x + 10 = 2 + 10 = 12 9(x + 2) = 9(2 + 2) = 36 (c) If x = – 4 If x = 2 54, 162, 486 108, 324, 942 (d) where x = – 4 Or x = 2 tn = arn – 1 = – 2(–3)n – 1 tn = arn – 1 = 4 × 3n – 1 (ii) (a) Here, x + 1, x – 5 and 1 are in GS then, x – 5 x + 1 = 1 x – 5 or, x2 – 10x + 25 = x + 1 or, x2 – 11x + 24 = 0 or, (x – 8) (x – 3) = x = {3, 8} (b) If x = 3, If x = 8, x + 1 = 3 + 1 = 4 x + 1 = 8 + 1 = 9 x – 5 = 3 – 5 = – 2 x – 5 = 8 – 5 = 3 1 = 1 1 = 1 (c) when, x = 3; – 1 2 , 1 4 and – 1 8 are three terms when, x = 8; 1 3 , 1 9 and 1 27 are three terms