According to the question, c 300 – 300 m hours = 2 hours
x x+5
⇒ 300 (x + 5) – 300x = 2 ⇒ 300 {(x + 5) – x} = 2
x (x + 5) x (x + 5)
⇒ 2x (x + 5) = 300 × 5 ⇒ 2x2 + 10x – 1500 = 0
⇒ x2 + 5x – 750 = 0 ⇒ x2 + 30x – 25x – 750 = 0
⇒ x(x + 30) – 25(x + 30) = 0 ⇒ (x – 25) (x + 30) = 0
⇒ x = 25 or x = – 30
⇒ x = 25 ( Speed cannot be negative)
Therefore, the usual speed of the train = 25 km/h.
11. A train travels at a certain average speed for a distance of 63 km and then travels at a distance
of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to
complete total journey, what is the original average speed? [CBSE 2018]
Sol. Let original average speed of the train = x km/h.
Distance travelled by train at this speed = 63 km
Time taken = 63 hours 63
x x+6
Now, if the average speed is increased by 6 km/h then time taken = hours
According to question,
63 + 72 =3 & 21 + 24 = 1
x x+6 x x+6
⇒ 21(x + 6) + 24x = x2 + 6x
⇒ x2 – 39x – 126 = 0
⇒ (x – 42)(x + 3) = 0
⇒ x = – 3 or x = 42
But x cannot be equal to –3 as speed cannot be negative.
Hence, original average speed of the train is 42 km/h.
12. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the
number, the digits interchange their places. Find the number.
Sol. Let the digit at tens place be x.
Then, digit at unit place = 18
x
\ Number = 10x + 18
x
18
and number obtained by interchanging the digits = 10 × x +x
According to question,
a10x + 18 k – 63 = 10 # 18 + x ⇒ a10x + 18 k – a10 # 18 + xk = 63
x x x x
⇒ 10x + 18 – 180 – x= 63 ⇒ 9x – 162 – 63 = 0
x x x
⇒ 9x2 – 63x – 162 = 0 ⇒ x2 – 7x – 18 = 0
⇒ x2 – 9x + 2x – 18 = 0 ⇒ x(x – 9) + 2(x – 9) = 0
⇒ (x – 9) (x + 2) = 0 ⇒ x = 9 or x = – 2
⇒ x = 9 [ a digit can never be negative]
Hence, the required number = 10 × 9 + 18 = 92.
9
Quadratic Equations 93
13. If twice the area of a smaller square is subtracted from the area of a larger square; the result is
14 cm2. However, if twice the area of the larger square is added to three times the area of the
smaller square, the result is 203 cm2. Determine the sides of the two squares.
Sol. Let the sides of the larger and smaller squares be x and y respectively. Then
x2 – 2y2 = 14 ...(i) and 2x2 + 3y2 = 203 ...(ii)
Operating (ii) –2 × (i), we get
2x2 + 3y2 – (2x2 – 4y2) = 203 – 2 × 14
⇒ 2x2 + 3y2 – 2x2 + 4y2 = 203 – 28
⇒ 7y2 = 175 ⇒ y2 = 25 ⇒ y = ±5
⇒ y = 5 [ Side cannot be negative]
By putting the value of y in equation (i), we get
x2 – 2 × 52 = 14 ⇒ x2 – 50 = 14 or x2 = 64
\ x = ± 8 or x = 8
\ Sides of the two squares are 8 cm and 5 cm.
14. If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would
have been 11 more than five times her actual age. What is her age now? [NCERT Exemplar]
Sol. Let the present age of Zeba be x years.
Age before 5 years = (x – 5) years
According to given condition,
⇒ (x – 5)2 = 5x + 11 ⇒ x 2 + 25 – 10x = 5x + 11
⇒ x2 – 15x + 14 = 0
⇒ x2 – 10x – 5x + 25 – 11 = 0
⇒ x(x – 14) –1(x – 14) = 0
⇒ x2 – 14x – x + 14 = 0
⇒ x – 1 = 0 or x – 14 = 0
⇒ (x – 1)(x – 14) = 0
x = 1 or x = 14
But present age cannot be 1 year.
\ Present age of Zeba is 14 years.
15. The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
[CBSE (F) 2014]
Sol. Let the two consecutive multiples of 7 be x and x +7.
x2 + (x + 7)2 = 637
⇒ x2 + x2 + 49 + 14x = 637 ⇒ 2x2 + 14x – 588 = 0
⇒ x2 + 7x – 294 = 0 ⇒ x2 + 21x – 14x – 294 = 0
⇒ x(x + 21) – 14(x + 21) = 0 ⇒ (x – 14) (x + 21) = 0
x = 14 or x = –21
\ The multiples are 14 and 21 or –21 and –14.
16. Solve for x: 1 + 2 = x 4 4 , x ! –1, – 2, – 4 [CBSE (AI) 2016]
x+1 x+2 +
Sol. x 1 1 + 2 = 4
+ x+2 x+4
⇒ x + 2 + 2 (x + 1) = 4 ⇒ (x + 4) (x + 2 + 2x + 2) = 4(x + 1)(x + 2)
(x + 1) (x + 2) x+4
⇒ (x + 4) (3x + 4) = 4(x2 + 3x + 2) ⇒ x2 – 4x – 8 = 0
⇒ x= 4! 16 + 32 = 4!4 3 =2!2 3
2 2
94 Xam idea Mathematics–X
17. Find the positive value(s) of k for which both quadratic equations x2 + kx + 64 = 0 and
x2 – 8x + k = 0 will have real roots.
[CBSE (F) 2016]
Sol. (i) For x2 + kx + 64 = 0 to have real roots
k2 – 4(1) (64) ≥ 0 i.e., k2 – 256 ≥ 0 ⇒ (k – 16) (k + 16) ≥ 0 ⇒ k ≤ –16 or k ≥ 16
(ii) For x2 – 8x + k = 0 to have real roots
(–8)2 – 4(k) ≥ 0 i.e., 64 – 4k ≥ 0 ⇒ k ≤ ± 16
For (i) and (ii) to hold simultaneously k = 16.
18. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same
point in 4 hours 30 minutes. Find the speed of the stream. [CBSE Delhi 2017]
Sol. Let the speed of stream be x km/h.
∴ Speed of boat in upstream = (15 – x) km/h
Speed of boat in downstream = (15 + x) km/h
According to question,
30 30 = 4 1 = 9
15 – x + 15 + x 2 2
⇒ 30 (15 + x + 15 – x) 9
(15 – x) (15 + x) = 2
⇒ 30 × 2 × 30 = 9(225 – x2) ⇒ 100 × 2 = 225 – x2
⇒ 200 = 225 – x2
⇒ x2 = 25 ⇒ x = ±5
⇒ x = 5 (Rejecting – 5)
∴ Speed of stream = 5 km/h
19. A motorboat whose speed in still water is 9 km/h, goes 15km downstream and comes back to the
same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream. [CBSE 2019 (30/3/3)]
Sol. Let speed of the stream be x km/h.
Speed of boat in upstream = (9 – x) km/h
Speed of boat in downstream = (9 + x) km/h
According to question,
15 + 15 =3 3
9– x 9+ x 4
⇒ 15 + 15 = 15
9– x 9+ x 4
⇒ 15 [9 + x + 9 – x] = 15
(9 – x) (9 + x) 4
⇒ 15 × 4 × 18 = 15(81 – x2)
⇒ 72 = 81 – x2
⇒ x2 = 9
⇒ x = 3, but x –3 (speed can’t be negative)
x = 3 ⇒ speed of stream = 3 km/h
20. Ram takes 6 days less than Bhagat to finish a piece of work. If both of them together can finish
the work in 4 days, in how many days Bhagat alone can finish the work. [CBSE Delhi 2017(C)]
Sol. Let Bhagat alone can do the work in x number of days.
∴ Ram takes (x – 6) number of days.
Quadratic Equations 95
Work done by Bhagat in 1 day = 1
x
1
Work done by Ram in 1 day = x–6
According to the question,
11 1
⇒ x + x–6 = 4
x–6+x = 1 ⇒ 4(2x – 6) = x(x – 6)
x (x – 6) 4
⇒ x2 – 14x + 24 = 0 ⇒ x2 – 12x – 2x + 24 = 0
⇒ x (x – 12) – 2 (x – 12) = 0
⇒ (x – 12) (x – 2) = 0 ⇒ Either x – 12 = 0 or x – 2 = 0
⇒ x = 12, as x ≠ 2
Bhagat alone can do the work in 12 days.
HOTS [Higher Order Thinking Skills]
1. One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had
gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total
number of camels.
Sol. Let x be the total number of camels. x
4
Then, number of camels in the forest =
Number of camels on mountains = 2 x
and number of camels on the bank of river = 15
Thus, total number of camels = x + 2 x + 15
4
Now by hypothesis, we have
x + 2 x + 15 = x ⇒ 3x – 8 x – 60 = 0
4
Let x = y , then x = y2
⇒ 3y2 – 8y – 60 = 0 ⇒ 3y2 – 18y + 10y – 60 = 0
⇒ 3y(y – 6) + 10(y – 6) = 0 ⇒ (3y + 10)(y – 6) = 0
⇒ y=6 or y = – 10
3
Now, y = – 10 ⇒ x = c– 10 2 = 100 ( x = y2)
3 3 9
m
But, the number of camels cannot be a fraction.
\ y = 6 ⇒ x = 62 = 36
Hence, the number of camels = 36
2. Solve the following quadratic equation:
9x2 – 9 (a + b) x + [2a2 + 5ab + 2b2] = 0 [CBSE (F) 2016]
Sol. Consider the equation 9x2 – 9 (a + b) x + [2a2 + 5ab + 2b2] = 0
Now comparing with Ax2 + Bx + C = 0, we get
A = 9, B = – 9 (a + b) and C = [2a2 + 5ab + 2b2]
Now discriminant,
96 Xam idea Mathematics–X
D = B2 – 4AC
= {– 9 (a + b)}2 – 4 × 9 (2a2 + 5ab + 2b2) = 92 (a + b)2 – 4 × 9 (2a2 + 5ab + 2b2)
= 9 {9 (a + b)2 – 4 (2a2 + 5ab + 2b2)} = 9 {9a2 + 9b2 + 18ab – 8a2 – 20ab – 8b2}
= 9 {a2 + b2 – 2ab} = 9 (a – b)2
Now using the quadratic formula,
x = – B! D , we get x = 9 (a + b) ! 9 (a – b) 2
⇒ 2A 2# 9
⇒
⇒ x = 9 (a + b) ! 3 (a – b) ⇒ x = 3 (a + b) ! (a – b)
2#9 6
x = (3a + 3b) + (a – b) and x = (3a + 3b) – (a – b)
6 6
x = (4a + 2b) and x = (2a + 4b)
6 6
⇒ x = 2a + b and x = a + 2b are required solutions.
3 3
3. Two taps running together can fill a tank in 3 1 hours. If one tap takes 3 hours more than the
13
other to fill the tank, then how much time will each tap take to fill the tank? [CBSE (AI) 2017]
Sol. Let, time taken by faster tap to fill the tank be x hours.
Therefore, time taken by slower tap to fill the tank = (x + 3) hours
Since the faster tap takes x hours to fill the tank.
\ Portion of the tank filled by the faster tap in one hour = 1
x
Portion of the tank filled by the slower tap in one hour = 1
x+3
Portion of the tank filled by the two tap together in one hour = 1 = 13
40 40
13
According to question,
⇒ 1 + 1 = 13 ⇒ x+3+x = 13
⇒ x x+3 40 x (x + 3) 40
40 (2x + 3) = 13x (x + 3) ⇒ 80x + 120 = 13x2 + 39x
⇒ 13x2 – 41x – 120 = 0 ⇒ 13x2 – 65x + 24x – 120 = 0
⇒ 13x (x – 5) + 24 (x – 5) = 0 ⇒ (x – 5) (13x + 24) = 0
Either x – 5 = 0 or 13x + 24 = 0
⇒ x = 5 or x = – 24
13
⇒ x = 5 [ x cannot be negative]
Hence, time taken by faster tap to fill the tank = x = 5 hours
and time taken by slower tap = x + 3 = 5 + 3 = 8 hours.
Quadratic Equations 97
4. In a rectangular park of dimensions 50 m × 40 m, a rectangular pond is constructed so that the
area of grass strip of uniform width surrounding the pond would be 1184 m2. Find the length
and breadth of the pond. [NCERT Exemplar, CBSE (F) 2017]
Sol. Let ABCD be rectangular lawn and EFGH be rectangular pond. Let x m be the width of grass
area, which is same around the pond.
Given, Length of lawn = 50 m
Width of lawn = 40 m
⇒ Length of pond = (50 – 2x) m
Breadth of pond = (40 – 2x) m
Also given, Fig. 4.1
Area of grass surrounding the pond = 1184m2
⇒ Area of rectangular lawn – area of pond = 1184 m2
⇒ 50 × 40 – {(50 – 2x) × (40 – 2x)} = 1184
⇒
⇒ 2000 – (2000 – 80x – 100x + 4x2) = 1184
⇒
⇒ 2000 – 2000 + 180x – 4x2 = 1184
4x2 – 180x + 1184 = 0 ⇒ x2 – 45x + 296 = 0
x2 – 37x – 8x + 296 = 0 ⇒ x(x – 37) – 8(x – 37) = 0
⇒ (x – 37) (x – 8) = 0 ⇒ x – 37 = 0 or x – 8 = 0
⇒ x = 37 or x = 8
x = 37 is not possible as in this case length of pond becomes 50 – 2 × 37 = – 24 (not possible)
Hence, x = 8 is acceptable
\ Length of pond = 50 – 2 × 8 = 34 m
Breadth of pond = 40 – 2 × 8 = 24 m
5. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side
is 30 metres more than the shorter side, find the sides of the field. [NCERT]
Sol. Let ABCD be the rectangular field. Let the shorter side BC of the rectangle = x metres.
According to question,
Diagonal of the rectangle, AC = (x + 60) metres
side of the rectangle, AB =(x + 30) metres
By Pythagoras theorem, AC2 = AB2 + BC2
\ (x + 60)2 = (x + 30)2 + x2
or x2 + 120x + 3600 = x2 + 60x + 900 +x2
\ (2x2 – x2) + (60x – 120x) + 900 – 3600 = 0
or x2 – 60x – 2700 = 0 Fig. 4.2
or (x – 90)(x + 30) = 0
\ Either x – 90 = 0 or x + 30 = 0
⇒ x = 90 or x = – 30 (But side cannot be negative)
So, the shorter side of rectangle = 30 m
and longer side of rectangle = 130 m
98 Xam idea Mathematics–X
PROFICIENCY EXERCISE
QQ Objective Type Questions: [1 mark each]
1. Choose and write the correct option in each of the following questions.
(i) The discriminant of the equation bx2 + ax + c = 0, b 0 is given by
(a) b2 – 4ac (b) a2 + 4bc (c) a2 – 4bc (d) b2 + 4ac
(ii) The roots of the equation x + 16 =10 are
x
(a) 4, 6 (b) 4, 4 (c) 4, 5 (d) 2, 8
(iii) If the roots of ax2 + bx + c = 0 are equal, then the value of c is
(a) –b (b) b (c) –b2 (d) b2
2a 2a 4a 4a
(iv) The value of k for which x = –2 is a root of the equation kx2 + x – 6 = 0 is
(a) –3 (b) –1 (c) –2 (d) 2
2
(v) If the discriminant of a quadratic equation is less than zero then it has
(a) equal roots (b) real roots (c) no real roots (d) can't be determined
2. Fill in the blanks.
(i) The value of k is _____________ if one root of the quadratic equation 6x2 – x – k = 0 is 2 .
3
(ii) A quadratic equation has _____________ roots if its discriminant is greater than zero.
(iii) Every quadratic equation has _____________ two roots.
(iv) The quadratic equation 3x2 – 7x + 4 = 0 has _____________ roots.
(v) If roots of a quadratic equation are equal, then its discriminant is equal to _____________.
QQ Very Short Answer Questions: [1 mark each]
3. If one root of 5x2 + 13x + k = 0 is the reciprocal of the other root, then find value of k.
[CBSE 2018 (C)]
4. For what values of k, the roots of the equation x2 + 4x + k = 0 are real? [CBSE 2019 (30/1/1)]
5. Find the value of k for which the roots of the equation 3x2 – 10x + k = 0 are reciprocal of each
other. [CBSE 2019 (30/1/1)]
6. Find the nature of roots of the quadratic equation 2x2 – 4x + 3 = 0. [CBSE 2019 (30/2/1)]
7. Find the nature of the roots of the quadratic equation 4x2 + 4 3 x + 3 = 0. [CBSE 2019 (30/2/3)]
8. For what values of k does the quadratic equation 4x2 – 12x – k = 0 have no real roots?
[CBSE 2019 (30/4/2)]
9. Find the value(s) of k for which the quadratic equation 3x2 + kx + 3 = 0 has real and equal roots.
[CBSE 2019 (30/5/1)]
10. For what values of 'a' the quadratic equation 9x2 – 3ax + 1 = 0 has equal roots?
[CBSE 2019 (C) (30/1/1)]
1
11. If one root of the quadratic equation 2x2 + 2x + k = 0 is – 3 , then find the value of k.
[CBSE 2019 (C) (30/1/1)]
12. What is the value of k for which 3 is a root of the equation kx2 – 7x + 3 = 0?
13. If the equation 16x2 + 6kx + 4 = 0 has equal roots, then what is the value of k?
Quadratic Equations 99
14. If ax2 + bx + c = 0 has equal roots, then what is b equal to?
15. What can be said about the quadratic equation x2 – 5x + 3 = 0?
16. What are the roots of the equation x2 – 9x + 20 = 0?
17. What is the value(s) of k for which the equation kx2 – kx + 1 = 0 has equal roots?
18. If the roots of the equation (a2 + b2) x2 – 2b(a + c)x + (b2 + c2) = 0 are equal, then what is the
value of b2?
19. What is the discriminant of the quadratic equation 3 3 x2 + 10x + 3 = 0 ?
20. Write a quadratic equation which has the product of two roots as 5.
21. If one ro ot of the quadra tic equa tion 6x2 – x – k = 0 is 32 , then find the value of k.
[CBSE (F) 2017]
QQ Short Answer Questions-I: [2 marks each]
State whether the following quadratic equations have two distinct real roots or not. Justify your answer.
(Q. 22 – 23)
22. 2x2 + x – 1 = 0
23. (x + 4)2 – 8x = 0 [NCERT Exemplar]
24. Write the condition to be satisfied for which equations ax2 + 2bx + c = 0 and bx2 – 2 ac x + b = 0
have equal roots.
25. If equation ax2 + bx + c = 0 has equal roots, then find ‘c’ in terms of ‘a’ and ‘b’.
QQ Short Answer Questions-II: [3 marks each]
26. Write all the values of p for which the quadratic equation x2 + px + 16 = 0 has equal roots. Find
the roots of the equation so obtained. [CBSE 2019, (30/2/1)]
27. Sum of the areas of two squares is 157 m2. If the sum of their perimeters is 68 m, find the sides
of the two squares. [CBSE 2019, (30/4/2)]
28. A motorboat whose speed is 18 km/h in still water takes one hour more to go 24 km upstream
than to return downstream to the same spot. Find the speed of the stream.
[CBSE 2018, 2019 (30/4/3), 2020 (30/5/1)]
29. Find the value of k for which the quadratic equation (k +1)x2 – 6(k + 1)x + 3(k + 9) = 0, k –1
has equal roots. [CBSE 2019 (C) (30/1/3)]
Find the roots of the following quadratic equations using the quadratic formula (Q. 30 – 31)
30. –3x2 + 5x + 12 = 0
31. x2 – 3 5 x + 10 = 0
32. Solve for x: 14 −1 = x 5 1 ; ≠ −3, −1 [CBSE Delhi 2014]
x+3 +
33. Solve for x: 3 – 1 = 2 1 ; x ! 1, x ! 1 [CBSE Delhi 2014]
x+1 2 3x – 3
Find the roots of the following quadratic equations by the factorisation method (Q. 34 – 37)
34. 3x2 + 5 5 x – 10 = 0
35. 2x2 + 5 x – 2 = 0 [NCERT Exemplar]
3 [CBSE Delhi 2015]
36. 4x2 – 4a2x + (a4 – b4) = 0
100 Xam idea Mathematics–X
37. x –1 + 2x + 1 = 2, x ! – 1 , 1 [CBSE (AI) 2017]
2x + 1 x –1 2
38. Find two consecutive positive integers, sum of whose squares is 925.
39. The sum of two numbers is 20 and the sum of their reciprocals is 5 . Find the numbers.
24
40. The difference of squares of two numbers is 88. If the larger number is 5 less than twice the
smaller number, then find the two numbers.
41. If one root of the quadratic equation 2x2 + kx – 6 = 0 is 2, find the value of k. Also find the other root.
42. If the roots of the equation (a2 + b2)x2 – 2(ac + bd) x + (c2 + d2) = 0 are equal, prove that ad = bc.
[CBSE 2019 (C), (30/1/1)]
OR
Show that if the roots of the following quadratic equation are equal, then ad = bc
x2(a2 + b2) + 2(ac + bd) x + (c2 + d2) = 0 [CBSE (AI) 2017, Delhi 2017 (C)]
43. Seven years ago, Varun’s age was five times the square of Swati’s age. Three years hence, Swati’s
age will be two fifth of Varun’s age. Find their present ages.
44. The area of a right angled triangle is 165 m2. Determine its base and altitude if the later exceeds
the former by 7 m.
45. Is it possible to design a rectangular park of perimeter 40 m and area 400 m2? If so, find its
length and breadth.
46. Determine the condition for the roots of the equation ax2 + bx + c = 0 to differ by 2.
47. Show that the equation 2(a2 + b2) x2 + 2(a + b)x + 1 = 0 has no real roots, when a ≠ b.
48. Find the roots of the equation: ax2 + a = a2x + x
49. If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2 – 2px + k = 0
has equal roots, find the value of k. [CBSE (F) 2014]
50. Find that non-zero value of k, for which the quadratic equation kx2 + 1 – 2(k – 1)x + x2 = 0 has
equal roots. Hence find the roots of the equation. [CBSE Delhi 2015]
QQ Long Answer Questions: [5 marks each]
[CBSE 2019 (30/4/3)]
51. Solve for x:
1 = 1 + 1 + 1 ; x ! 0, x ! –2a – b , a, b ! 0
2a + b + 2x 2a b 2x 2
52. The sum of the areas of two squares is 640 m2. If the difference of their perimeters is 64 m, find
the sides of the squares. [CBSE 2019 (30/4/3)]
53. Solve for x:
a +1b + x = 1 + 1 + 1 ; a ! b ! 0, x ! 0, x ! – (a + b) [CBSE 2019 (30/5/1)]
a b x
54. Solve the following equation for x: [CBSE 2019(C)(30/1/1)]
1 + 2 = x 7 5 , x ! –1, –2, –5
x+ 1 x+2 +
55. A motorboat whose speed is 18 km/h in still water takes 1 h 30 minutes more to go 36 km
upstream than to return downstream to the same spot. Find the speed of the stream.
[CBSE 2019(C)(30/1/3)]
Quadratic Equations 101
Find whether the following equations have real roots. If yes find them. (Q. 56 to 60)
56. 5x2 – 6x – 2 = 0
57. 8x2 + 2x – 3 = 0
58. 1 + 1 = 1, x ! 3 , 5 [NCERT Exemplar]
2x – 3 x–5 2
59. 3x2 – 2x + 2 = 0
60. 4x2 – 1 x + 1 =0
5 32
61. Solve for x: x – 3 + x –5 = 10 ; x ! 4, 6 [CBSE (AI) 2014]
x – 4 x–6 3 [CBSE (AI) 2014]
62. Solve for x: x–2 + x–4 = 10 ; x ≠ 3, 5
x–3 x–5 3
63. The sum of the squares of two consecutive even numbers is 340. Find the numbers. [CBSE (F) 2014]
64. Solve for x: x 2 1 + 3 2h = 23 , x ≠ 0, –1, 2 [CBSE Delhi 2015]
65. + 2^x – 5x 5 and the difference of their
5
The difference of two natural numbers is reciprocals is 14 . Find the
numbers. [CBSE Delhi 2014]
66. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given
number. [NCERT Exemplar]
67. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When
Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the
present age of Nisha. Find the present ages of both Asha and Nisha. [NCERT Exemplar]
68. There is a square field whose side is 44 m. A square flower bed is prepared in its centre leaving
a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the
path at ™2.75 and ™1.50 per m2 respectively, is ™4904. Find the width of gravel path.
69. At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found
t2
to be 3 minutes less than 4 minutes. Find t. [NCERT Exemplar]
70. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/h more, it
would have taken 30 minutes less for the journey. Find the original speed of the train.
71. In a class test, the sum of the marks obtained by Puneet in Mathematics and Science is 28. Had
he got 3 marks more in Mathematics and 4 marks less in Science, the product of their marks,
would have been 180. Find his marks in two subjects.
72. If the list price of a toy is reduced by ™2, a person can buy 2 toys more for ™360. Find the original
price of the toy.
73. Solve for x: 3 + 4 = 29 1 ; x ! 1, – 1, 1 [CBSE Delhi 2015]
x+1 x –1 4x – 4
74. Solve for x: 4x2 + 4bx – (a2 – b2) = 0 [CBSE (F) 2017]
[CBSE Delhi 2016]
75. Find x in terms of a, b and c: a + b = 2c c , x ≠ a, b, c.
x–a x–b x– [CBSE (AI) 2017]
76. Solve for x:
x 1 1 + 3 1 = x 5 4 , x ! –1, – 1 , –4
+ 5x + + 5
102 Xam idea Mathematics–X
77. Solve for x:
xx +– 3 – 1– x = 17 ; x ! 0, 2 [CBSE (AI) 2017]
2 x 4
78. A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of
slower train is 10 km/h less than that of faster train, find the speeds of two trains.
[CBSE 2018(C)]
79. Two water taps together can fill a tank in 1 7 hours. The tap with longer diameter takes 2 hours
8
less than the tap with smaller one to fill the tank separately. Find the time in which each tap can
fill the tank separately. [CBSE 2019(30/1/1)]
80. The total cost of a certain length of a piece of cloth is `200. If the piece was 5 m longer and each
metre of cloth costs `2 less, the cost of the piece would have remained unchanged. How long is
the piece and what is its original rate per metre? [CBSE 2019(30/2/2)]
81. A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km
away on time, it has to increase its speed by 250 km/h from its usual speed. Find the usual speed
of the plane. [CBSE 2019(30/4/2)]
82. Find the dimensions of a rectangular park whose perimeter is 60 m and area is 200 m2.
[CBSE 2019(30/4/2)]
Answers
1. (i) (c) (ii) (d) (iii) (d) (iv) (d) (v) (c)
2. (i) 2 (ii) real (iii) atmost (iv) real (v) zero
3. k = 5 4. k ≤ 4 5. k = 3 6. No real root 7. Real and equal 8. k < –9
9. k = 6 10. a = 2 11. k = 4 12. 2 13. ± 8 14. ±2 ac
9 17. k = 0, 4 3 19. 64
15. It has two distinct real roots 16. 4 and 5 18. ac
20. x2 – 2 5x + 5 = 0 21. k = 2 22. Yes, D > 0 23. No, D < 0 24. b2 = ac
25. c = b2 26. p = – 4, 4 27. Length of side of squares are 6 m and 11 m
4a
–4
28. Speed of stream 6 km/h 29. k = 3, 30. 3 , 3 31. 5, 2 5 32. x = 1 and 4
33. x = 1 and 3 34. x = 5 , −2 5 35. x = –3 , 2
3 2 3
a2 + b2 a2 b2
36. x = 2 , – 37. –2, –2 38. 22 and 21 39. 8 and 12 40. 9 and 13
41. 2 –3
2
k = –1, other root =
43. Swati’s age = 9 years; Varun’s age = 27 years 44. Base=15 m, Altitude=22 cm 45. No
46. b2 – 4ac = 4a2 48. a, 1 49. k = 1 50. k = 3, x = 1 , 1
a 2 2
b
51. x=– a or – 2 52. Length of sides of squares are 24 m and 8 m
53. x = – a, x = – b 54. x =1, x= –3 55. Speed of stream = 6 km/h
2
56. Yes; 3± 19 57. Yes; −3 , 1 58. Yes; 8±3 2
5 4 2
2
Quadratic Equations 103
59. No 60. No real roots 61. 9 and 7 62. 7 and 6 63. 12 and 14
2 2
−23
64. 4 and 11 65. 7 and 2 66. 12
67. Nisha’s age is 5 years, Asha’s age is 27 years 68. 2 m 69. 14
70. 45 km/h 71. Mathematics: 9 and Science: 19 or Mathematics: 12 and Science: 16
a–b – a–b
72. ™20 73. x = –7, 4 74. 2, 2 7 5. x = 2ab – ac – bc , 0 76. –11 , 1
a+ b– 2c 17
–2
77. 4, 9 78. Speed of faster train = 50 km/h and speed of slower train = 40 km/h
79. 5 hours, 3 hours 80. l = 20 m, `10 per metre
81. Speed of plane = 750 km/h 82. Length = 20 m, breath = 10 m
SELF-ASSESSMENT TEST
Time allowed: 1 hour Max. marks: 40
Section A
1. Choose and write the correct option in the following questions. (4 × 1 = 4)
(i) The smallest positive value of k for which the equation x2 + kx + 9 = 0 has real roots is
(a) – 6 (b) 6 (c) – 36 (d) – 3
(ii) If ax2 + bx + c = 0 has equal roots, then a is equal to
(a) b2 (b) – b2 (c) b (d) –b
4c 4c 2c 2c
(iii) The value of k for which x = – 2 is a root of the quadratic equation kx2 + x – 6 = 0 is
(a) – 1 (b) – 2 (c) 2 (d) –3
2
(iv) If px2 + 3x + q = 0 has two roots x = – 1 and x = – 2, the value of q – p is
(a) – 1 (b) 1 (c) 2 (d) – 2
2. Fill in the blanks. (3 × 1 = 3)
(i) The degree of polynomial 2x2 + bx + c is _______________.
(ii) The equation 16x2 + 6kx + 4 = 0 has equal roots, then the value of k is _______________ .
(iii) The roots of the equation x2 – 9x + 20 = 0 are _______________ .
3. Solve the following questions. (3 × 1 = 3)
(i) What is the condition that the equation (a2 + b2) x2 + 2 (ac + bd) x + c2 + d2 = 0 has no real
roots?
(ii) What are the roots of the quadratic equation 6x2 – x – k = 0 if value of k is equal to 2?
(iii) Check whether 4 + 1 –5 = 0 is a quadratic equation or not? If it is quadratic then write
x2 x
in which variable it is quadratic?
Section B
QQ Solve the following questions. (3 × 2 = 6)
4. Solve for x: x+3 = 3x – 7 , x ! –2, 3 [CBSE Delhi 2017 (C)]
x+2 2x – 3 2
5. Find the value of k such that the equation (k – 12) x2 + 2(k – 12) x + 2 = 0 has equal roots.
[CBSE Delhi 2017 (C)]
6. Solve for x: 3 x2 + 10x + 7 3 = 0 [CBSE (F) 2017]
104 Xam idea Mathematics–X
QQ Solve the following questions. (3 × 3 = 9)
7. The difference of two natural numbers is 3 and the difference of their reciprocals is 3 . Find
28
the numbers. [CBSE Delhi 2014]
8. Find the value of c for which the quadratic equation 4x2 – 2(c + 1)x + (c + 1) = 0 has equal roots,
which are real. [CBSE Delhi 2017 (C)]
9. Solve for x: 1 = 1 + 1 + 1 ; a +b+x ≠ 0, a, b, x ≠ 0 [CBSE (F) 2017, 2018 (C)]
a+b+x a b x
QQ Solve the following questions. (3 × 5 = 15)
10. Three hundred apples are distributed equally among a certain number of students. Had there
been 10 more students, each would have received one apple less. Find the number of students.
11. Two water taps together can fill a tank in 9 hours 36 minutes. The tap of larger diameter takes
8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can
separately fill the tank. [CBSE (F) 2016]
12. A motorboat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than
to return downstream to the same spot. Find the speed of the stream. [CBSE (AI) 2016]
Answers
1. (i) (b) (ii) (a) (iii) (c) (iv) (b)
2. (i) 2 (ii) ! 8 (iii) 4 and 5
3
3. (i) ad < bc (ii) 2 and –1 (iii) Yes, 1
3 2 x
4. 5, –1 5. k = 14 6. –7 , – 3 7. 7 and 4 8. c = –1, 3
3 11. 24 hours, 16 hours
9. x = –a, –b 10. 50 students
12. 8 km/h
zzz
Quadratic Equations 105
5 Arithmetic
Progressions
BASIC CONCEPTS – A FLOW CHART
106 Xam idea Mathematics–X
MORE POINTS TO REMEMBER
Selection of term in an AP: In some particular problems we require certain number of
terms in AP. For convenience we adopt following pattern
Number of terms Terms Common difference
3 a – d, a, a + d d
4 a – 3d, a – d, a + d, a + 3d 2d
5 a – 2d, a – d, a, a + d, a + 2d d
6 a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d 2d
A sequence a1, a2, a3.......is an AP, if an+1 – an is independent of n.
For example, 2, 5, 8, 11, .......is an AP because
an + 1 – an = {2+(n + 1 – 1)3} – {2 + (n – 1)3}
= {2 + 3n – 2 – 3n + 3} = 3 (independent of n)
A sequence a1, a2, a3.........., an, ......... is an AP, if and only if nth term an is a linear expansion
of n and the co-efficient of n is common difference of the AP.
For example, if a be the first term and d the common difference then
an = a + (n – 1) d
= a + nd – d = dn + (a – d)
Multiple Choice Questions [1 mark]
Choose and write the correct option in the following questions.
1. The first term and the common difference for the AP 4 , 1 , −2 , −5 , .... are respectively
3 3 3 3
(a) 4 ,1 (b) 4 , –1 (c) 1, 4 (d) –1, 4
3 3 3 3
2. The 11th and 13th terms of an AP are 35 and 41 respectively, its common difference is
(a) 38 (b) 32 (c) 6 (d) 3
3. The next term of the AP 8, 18, 32, .... is
(a) 5 2 (b) 5 3 (c) 3 3 (d) 5 3
4. Which term of the AP 18, 23, 28, 33, ... is 98?
(a) 15th (b) 16th (c) 17th (d) 18th
5. The famous mathematician associated with finding the sum of the first 100 natural numbers
is [NCERT Exemplar]
(a) Pythagoras (b) Newton (c) Gauss (d) Euclid
6. The 16th term of the AP 15, 25 , 10, 15 , 5, .... is
2 2
(a) 45 (b) –45 (c) 105 (d) – 105
2 2 2 2
7. The 6th term from the end of the AP 5, 2, –1, – 4, ..., –31 is
(a) –16 (b) –19 (c) –22 (d) –25
8. The list of numbers –10, –6, –2, 2, ... is [NCERT Exemplar]
(a) an AP with d = –16 (b) an AP with d = 4
(c) an AP with d = –4 (d) not an AP
Arithmetic Progressions 107
9. Which term of the AP 21, 42, 63, 84, ... is 210? [NCERT Exemplar]
(d) 12th
(a) 9th (b) 10th (c) 11th
10. The sum of first 16 terms of the AP 10, 6, 2, ... is [NCERT Exemplar]
(d) –400
(a) –320 (b) 320 (c) –352
11. In an AP if a = 1, an = 20 and Sn = 399, then n is [NCERT Exemplar]
(c) 38 (d) 42
(a) 19 (b) 21
12. The sum of first five multiples of 3 is [NCERT Exemplar]
(d) 75
(a) 45 (b) 55 (c) 65
13. The 4th term from the end of the AP –11, –8, –5, ..., 49 is [NCERT Exemplar]
(d) 58
(a) 37 (b) 40 (c) 43
14. Two APs have same common difference. The first term of one of these is –1 and that of the
other is – 8. Then the difference between their 4th term is [NCERT Exemplar]
(a) –1 (b) –8 (c) 7 (d) –9
15. If (p – 1), (p + 3), (3p – 1) are in AP, then p is equal to
(a) 4 (b) – 4 (c) 2 (d) –2
16. The 21st term of the AP whose first two terms are –3 and 4, is [NCERT Exemplar]
(d) –143
(a) 17 (b) 137 (c) 143
(d) 3n + 4
17. If the sum of n terms of an AP is 2n2 + 5n, then its nth term is
(a) 4n – 3 (b) 3n – 4 (c) 4n + 3
18. If 18, a, b, –3 are in AP, then a + b is equal to
(a) 19 (b) 7 (c) 11 (d) 15
19. If the sum of p terms of an AP is q and the sum of q terms is p, then the sum of p + q terms will
be
(a) 0 (b) p – q (c) p + q (d) – (p + q)
20. The first four terms of an AP, whose first term is –2 and the common difference is –2 are
[NCERT Exemplar]
(a) –2, 0, 2, 4 (b) –2, 4, –8, 16 (c) –2, –4, –6, –8 (d) –2, –4, –8, –16
Answers
1. (b) 2. (d) 3. (a) 4. (c) 5. (c) 6. (b)
7. (a) 8. (b) 9. (b) 10. (a) 11. (c) 12. (a)
13. (b) 14. (c) 15. (a) 16. (b) 17. (c) 18. (d)
19. (d) 20. (c)
Fill in the Blanks [1 mark]
Complete the following statements with appropriate word(s) in the blank space(s).
1. An arithmetic progression is a list of numbers in which each term is obtained by _______________
a fixed number to the preceding term except the first term.
2. In an AP, the letter d is generally used to denote the _______________.
3. The nth term of an AP is always a _______________ expression.
108 Xam idea Mathematics–X
4. If Sn denotes the sum of n term of an AP, then S12 – S11 is the _______________ term of the AP.
5. If a and d are respectively the first term and the common difference of an AP, a + 10d, denotes
the _______________ term of the AP.
6. If l and d are respectively the last term and the common difference of an AP, then l – 9d denotes
the _______________ term from end of the AP.
7. In the sequence 5, 6, 7, 8 difference between two consecutive terms is _________________.
8. If a, b, c are in AP, then the relation between a, b and c is _________________.
9. The sum of the AP, 1 + 2 + ..... + 10 is _________________.
10. The nth term of an AP whose first term is a and common difference is d is _______________.
11. The sum of three numbers in AP is 30. If the greatest is 13 then, its common difference is
_________________.
12. The sum of first n natural numbers is _________________.
13. Natural numbers up to 300 which are divisible by 17 in number are _________________.
14. If 2x, (x + 10) and (3x + 2) are in AP then x is _________________.
15. The sum of all even numbers between 100 and 200 will be _________________.
Answers
1. adding 2. common difference 3. linear 4. twelfth 5. eleventh
6. tenth 9. 55 10. a + (n – 1)d
7. 1 8. 2b = a + c or b = a+c
11. 3 2 15. 7350
n]n +
12. 2 1g 13. 17 14. 6
Very Short Answer Questions [1 mark]
1. Which of the following can be the nth term of an AP?
4n + 3, 3n2 + 5, n3 + 1 give reason.
Sol. 4n + 3 because nth term of an AP can only be a linear relation in n as an = a + (n – 1)d.
2. Is 144 a term of the AP 3, 7, 11, ...? Justify your answer.
Sol. No, because here a = 3 an odd number and d = 4 which is even. So, sum of odd and even must
be odd whereas 144 is an even number.
3. The first term of an AP is p and its common difference is q. Find its 10th term.
Sol. a10 = a + 9d = p + 9q 3+ n
4
4. If nth term of an AP is , find its 8th term.
Sol. an = 3 + n ; So a8 = 3+8 = 11
4 4 4
5. For what value of p are 2p + 1, 13, 5p – 3, three consecutive terms of AP? [CBSE (AI) 2009]
Sol. Since 2p + 1, 13, 5p – 3 are in AP.
\ Second term – First term = Third term – Second term
13 – (2p + 1) = 5p – 3 – 13
13 – 2p – 1 = 5p – 16
⇒ 12 – 2p = 5p – 16
⇒ – 7p = – 28
⇒ p= 4
Arithmetic Progressions 109
6. Write the common difference of the AP 3, 12, 27, 48, ... [CBSE 2019 (30/5/2)]
Sol. Given AP, 3 , 12, 27, 48, ...
⇒ 3, 2 3, 3 3, 4 3, ...
∴ Common difference = 2 3 – 3 = 3
⇒ d = 3
7. If 7 times the 7th term of an AP is equal to 11 times its 11th term, then find its 18th term.
[CBSE (F) 2017]
Sol. Given, 7a7 = 11a11
⇒ 7(a + 6d) = 11(a + 10d) or 7a + 42d = 11a + 110d
⇒ 4a + 68d = 0 or a + 17d = 0
Now, a18 = a + 17d = 0
8. Find the 9th term from the end (towards the first term) of the AP 5, 9, 13, ..., 185.
[CBSE (Delhi) 2016]
Sol. l = 185, d = 4
l9 = l – (n – 1) d = 185 – 8 × 4 = 153
Short Answer Questions-I [2 marks]
1. In which of the following situations, does the list of numbers involved to make an AP? If yes,
give reason.
(i) The cost of digging a well after every metre of digging, when it costs ™150 for the first
metre and rises by ™50 for each subsequent metre.
(ii) The amount of money in the account every year, when ™10,000 is deposited at simple
interest at 8% per annum.
Sol. (i) The numbers involved are 150, 200, 250, 300, ...
Here 200 –150 = 250 – 200 = 300 – 250 and so on
\ It forms an AP with a = 150, d = 50
(ii) The numbers involved are 10,800, 11,600, 12,400, ...
which forms an AP with a = 10,800 and d = 800.
2. If the sum of the first p terms of an AP is ap2 + bp, find its common difference.
Sol. ap = Sp – Sp–1 = (ap2 + bp) –[a(p – 1)2 + b(p – 1)]
= ap2 + bp – (ap2 + a – 2ap + bp – b)
= ap2 + bp – ap2 – a + 2ap – bp + b = 2ap + b – a
\ a1 = 2a + b – a = a + b and a2 = 4a + b – a = 3a + b
⇒ d = a2 – a1 = (3a + b) – (a + b) = 2a
3. The first and the last term of an AP are 5 and 45 respectively. If the sum of all its terms is 400,
find its common difference. [CBSE Delhi 2014]
Sol. Let the first term be ‘a’ and common difference be ‘d’.
Given, a = 5, Tn = 45, Sn = 400
Tn = a + (n – 1)d ⇒ 45 = 5 + (n – 1)d
⇒ (n – 1) d = 40 ...(i)
Sn = n (a + Tn) ⇒ 400 = n (5 + 45)
2 2
⇒ n = 2 × 8 = 16 40 8
Substituting the value of n in (i) 15 3
(16 – 1)d = 40 ⇒ d= =
110 Xam idea Mathematics–X
4. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
[CBSE (AI) 2014]
Sol. Natural numbers between 101 and 999 divisible by both 2 and 5 are 110, 120, … 990.
So, a1 = 110, d = 10, an = 990
We know, an = a1 + (n –1)d
990 = 110 + (n – 1) 10
(n – 1) = 990 – 110 ⇒ n = 88 + 1 = 89
10
5. Find how many integers between 200 and 500 are divisible by 8. [CBSE Delhi 2017]
Sol. AP formed is 208, 216, 224, …, 496
Here, an = 496, a = 208, d = 8
⇒ an = a + (n – 1) d ⇒ 208 + (n – 1) × 8 = 496
n – 1 = 36
8 (n – 1) = 288 ⇒
⇒ n = 37
6. If Sn, the sum of the first n terms of an AP is given by Sn = 2n2+n, then find its nth term.
[CBSE 2019 (30/5/1)]
Sol. We have in an AP
Sn = 2n2 + n
∴ Sn–1 = 2(n – 1)2 + (n – 1) = 2n2 – 4n + 2 + n – 1
⇒ Sn–1 = 2n2 – 3n + 1
∴ Its nth term, tn = Sn – Sn–1
= 2n2 + n – 2n2 +3n – 1
⇒ tn = 4n – 1
7. How many terms of the AP 18, 16, 14, .... be taken so that their sum is zero? [CBSE Delhi 2016]
Sol. Here, a = 18, d = –2, Sn = 0
Therefore, n [36 + (n – 1) (– 2)] = 0
2
⇒ n(36 – 2n + 2) = 0 ⇒ n(38 – 2n) = 0 ⇒ n = 19
8. The 4th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term.
[CBSE (AI) 2016]
Sol. a4 = 0 (Given)
⇒ a + 3d = 0 ⇒ a = –3d
a25 = a + 24d = – 3d + 24d = 21d
3a11 = 3(a + 10d) = 3(7d) = 21d
\ a25 = 3a11 Hence proved.
9. If the ratio of sum of the first m and n terms of an AP is m2 : n2, show that the ratio of its mth
and nth terms is (2m – 1) : (2n – 1). [CBSE (F) 2016, Delhi 2017]
Sm m2 m (2a +(m – 1) d)
Sn n2 2 (2a +(n – 1) d)
Sol. = = n
2
2a +(m – 1) d
⇒ m = 2a +(n – 1) d ⇒ 2am + mnd – md = 2an + mnd – nd
n
⇒ a(2m – 2n) = d(m – n) ⇒ 2a = d
aamn a+(m – 1) d a+ 2 (m – 1) a
= a+(n – 1) d = a+ 2 (n – 1) a = 2m –1 Hence proved.
2n –1
Arithmetic Progressions 111
10. What is the common difference of an AP in which a21 – a7 = 84? [CBSE (AI) 2017]
d=6
Sol. Given: a21 – a7 = 84
⇒ (a + 20d) – (a + 6d) = 84 ⇒ 14d = 84 ⇒
11. For what value of n, are the nth terms of two APs 63, 65, 67,... and 3, 10, 17,... equal?
[CBSE (AI) 2017]
Sol. Let nth terms for two given series be an and a′n
According to question,
an = a′n ⇒ a + (n – 1)d = a′ + (n – 1)d′
⇒ 63 + (n – 1)2 = 3 + (n – 1)7
⇒ 5n = 65 ⇒ n = 13.
12. If the 17th term of an AP exceeds its 10th term by 7, find the common difference.
[CBSE 2019 (30/5/1)]
Sol. According to question,
a17 = a10 + 7
⇒ a + (17 – 1) d = a + (10 – 1) d + 7
⇒ a + 16d = a + 9d + 7
⇒ 16d – 9d = 7
⇒ 7d = 7 ⇒ d = 1
\ Common difference = 1
13. Find the 25th term of the AP –5, –5 , 0, 5 .... [CBSE 2009]
2 2
–5 5 n=5+1=6
Sol. Here, a = –5, d = 2 – (–5) = 2
[3 marks]
We know that, a25 = a + (25–1) d
= (–5) + 24 d 5 n
2
= –5 + 60 = 55
14. If in an AP, a = 15, d = –3 and an = 0, then find the value of n.
Sol. In an AP it is given that
a = 15, d = –3 and an = 0
Now, an = a + (n – 1)d
0 = 15 + (n – 1) × (–3) 15
3
⇒ 3(n – 1) = 15 ⇒ n–1= =5 ⇒
∴ n = 6
Short Answer Questions-II
1. Which term of the AP 3, 8, 13, 18, ... , is 78? [NCERT]
Sol. Let an be the required term and we have given AP 3, 8, 13, 18, .....
Here, a = 3, d = 8 – 3 = 5 and an = 78
Now, an = a + (n – 1)d ⇒ 78 = 3 + (n – 1) × 5
⇒ 78 – 3 = (n – 1) × 5 ⇒ 75 = (n – 1) × 5
⇒ 75 = n – 1 ⇒ 15 = n – 1 ⇒ n = 15 + 1 = 16
5
Hence, 16th term of given AP is 78.
112 Xam idea Mathematics–X
2. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. [NCERT]
Sol. Let the first term be a and common difference be d. ...(i)
...(ii)
Now, we have
a11 = 38 ⇒ a + (11 – 1)d = 38
⇒ a + 10d = 38
and a16 = 73 ⇒ a + (16 – 1)d = 73
⇒ a + 15d = 73
Now subtracting (ii) from (i), we have
Now, a + 10d = 38
– a +– 15d =–73
–5d = –35 or 5d = 35
\ d = 35 =7
5
Putting the value of d in equation (i), we have
a + 10 × 7 = 38 ⇒ a + 70 = 38
⇒ a = 38 – 70 ⇒ a = – 32
We have, a = – 32 and d = 7
Therefore, a31 = a + (31 – 1)d
⇒ a31 = a + 30d = (–32) + 30 × 7 = – 32 + 210
⇒ a31 = 178
3. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th
term. [NCERT]
Sol. Let a be the first term and d be the common difference.
Since, given AP consists of 50 terms, so n = 50
a3 = 12 ⇒ a + 2d = 12 ...(i)
...(ii)
Also, a50 = 106 ⇒ a + 49d = 106
Subtracting (i) from (ii), we have 94
47
47d = 94 ⇒ d = =2
Putting the value of d in equation (i), we have
a + 2 × 2 = 12 ⇒ a = 12 – 4 = 8
Here, a = 8, d = 2
\ 29th term is given by
a29 = a + (29 – 1)d = 8 + 28 × 2
⇒
a29 = 8 + 56 ⇒ a29 = 64
4. Which term of the arithmetic progression 3, 15, 27, 39 .... will be 120 more than its 21st
term? [CBSE 2019 (30/1/2)]
Sol. We have, a = 3 and d = 12
\ a21 = a + 20d = 3 + 20 × 12 = 243
Let nth term of the given AP be 120 more than its 21st term. Then,
an = 120 + a21
\ 3 + (n – 1)d = 120 + 243
⇒ 3 + 12(n – 1) = 363 ⇒ 12(n – 1) = 360
⇒ n – 1 = 30 ⇒ n = 31
Hence, 31st term of the given AP is 120 more than its 21st term.
Arithmetic Progressions 113
So5l.. HW ehriechdt= erm19o14f t–h 2e0p=rog74r7 es–si2o0n= 20–43, 1,9 1a4 ,=182102 , 17 3 , ... is the first negative term?
4 [CBSE (AI) 2017]
Let the nth term be first negative term.
∴ a + (n – 1)d < 0
So, 20 + (n – 1)c –3 m < 0 ⇒ 80 – 3n + 3 < 0
4
⇒ 3n > 83
⇒ n > 27 2
3
Hence 28th term is first negative term.
6. In an AP, given l = 28, S = 144 and there are total 9 terms. Find a. [NCERT]
Sol. We have, l = 28, S = 144 and n = 9
Now, l = an = 28 Aliter
28 = a + (n – 1) d ⇒ 28 = a + (9 – 1) d n
2
⇒ 28 = a + 8d ...(i) S = ]a + lg
and S = 144 9 ` 144 = 9 "a + 28,
n 2 2
⇒ 144 = 2 [2a +(n – 1) d] ⇒ 144 = [2a +(9 – 1) d]
144 ×2
144 # 2 & a + 28 = 9
9
= 2a + 8d ⇒ 32 = 2a + 8d = 32
⇒ 16 = a + 4d ...(ii) & a = 32 – 28
Now, subtracting equation (ii) from (i), we get & a=4
4d = 12 or d = 3
Putting the value of d in equation (i), we have
a + 8 × 3 = 28
⇒ a + 24 = 28 ⇒ a = 28 – 24 = 4
\ a = 4.
7. How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636?
[NCERT, CBSE (AI) 2017]
Sol. Let sum of n terms be 636.
Sn = 636, a = 9, d = 17 – 9 = 8
⇒ n [2a + (n – 1) d] = 636 ⇒ n [2 × 9 + (n – 1) × 8] = 636
2 2
n
⇒ 2 × 2[9 + (n – 1) × 4] = 636 ⇒ n[9 + 4n – 4] = 636
⇒ n[5 + 4n] = 636 ⇒ 5n + 4n2 = 636
⇒ 4n2 + 5n – 636 = 0
\ n= – 5! (5)2 – 4 # 4 # (– 636) = – 5! 25 +10176
2#4 8
= – 5! 10201 = – 5 !101 = 96 , –106 =12, – 53
8 8 8 8 4
But n≠ – 53 So, n = 12
4
Thus, the sum of 12 terms of given AP is 636.
114 Xam idea Mathematics–X
8. How many terms of the series 54, 51, 48 ....... be taken so that, their sum is 513? Explain the
double answer.
Sol. Clearly, the given sequence is an AP with first term a = 54 and common difference d = – 3. Let
the sum of n terms be 513. Then,
Sn = 513
⇒
n {2a + (n – 1) d} = 513 ⇒ n [108 + (n – 1) × (– 3)] = 513
2 2
⇒ n[108 – 3n + 3] = 1026 ⇒ –3n2 + 111n = 1026
⇒ n2 – 37n + 342 = 0 ⇒ (n – 18) (n – 19) = 0
⇒ n = 18 or 19
Here, the common difference is negative. So, 19th term is given by
a19 = 54 + (19 – 1) × (– 3) = 0
Thus, the sum of 18 terms as well as that of 19 terms is 513.
9. The first term, common difference and last term of an AP are 12, 6 and 252 respectively. Find
the sum of all terms of this AP.
Sol. We have, a = 12, d = 6 and l = 252
Now, l = 252 ⇒ an = 252
⇒ l = a + (n – 1) d ⇒ 252 = 12 + (n – 1) × 6
⇒ 240 = (n – 1) × 6 ⇒ n – 1 = 40 or n = 41
Thus, Sn = n (a + l)
2
⇒ S41 = 41 (12 + 252) = 41 (264) = 41 × 132 = 5412
2 2
10. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n
terms. [CBSE Delhi 2016, (F) 2017]
Sol. We have, S7 = 49 2
7 7
⇒ 49 = 2 [2a + (7 – 1) × d] ⇒ 49 × = 2a + 6d
⇒ 14 = 2a + 6d ⇒ a + 3d = 7 ...(i)
and S17 = 289
⇒ 289 = 17 [2a + (17 – 1) d] ⇒ 2a + 16d = 289 # 2 = 34
⇒ 2 17
a + 8d = 17 ...(ii)
Now, subtracting equation (i) from (ii), we have
5d = 10 ⇒ d=2
Putting the value of d in equation (i), we have
a + 3 × 2 = 7 ⇒ a=7–6=1
Here, a = 1 and d = 2
Now, Sn = n [2a + (n – 1) d]
2
= n [2 × 1 + (n – 1) × 2] = n [2 + 2n – 2] = n × 2n = n2
2 2 2
Arithmetic Progressions 115
11. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms
and the common difference. [CBSE (AI) 2017]
Sol. We have, a = 5, l = 45 and Sn = 400
⇒ an = 45 ⇒ a + (n – 1) d = 45
⇒ 5 + (n – 1) × d = 45 ⇒ (n – 1) d = 40 ...(i)
n
Again Sn = 400 ⇒ 2 [2a + (n – 1) d] = 400
⇒ n [2 × 5 + (n – 1)d] = 400
2
n [10 + 40] = 400 (Using equation (i))
2
⇒ n × 50 = 400 ⇒ n= 400 =16
2 25
Now, putting the value of n in equation (i), we have
(16 – 1) d = 40 ⇒ 15d = 40
⇒ d= 40 = 8
15 3
8
Hence, number of terms is 16 and common difference is 3 .
12. The sum of the 5th and the 9th terms of an AP is 30. If its 25th term is three times its 8th term,
find the AP. [CBSE (AI) 2014]
Sol. According to question,
a5 + a9 = 30
⇒ (a + 4d) + (a + 8d) = 30
⇒ 2a + 12d = 30 ⇒ a + 6d = 15
⇒ a = 15 – 6d ...(i)
a25 = 3a8 ⇒ a + 24d = 3(a + 7d)
a + 24d = 3a + 21d ⇒ 2a = 3d
Putting the value of a form (i), we have
2 (15 – 6d) = 3d ⇒ 30 – 12d = 3d
⇒ 15d = 30 ⇒ d=2
So, a = 15 – 6 × 2 = 15 – 12 [From equation (i)]
⇒ a = 3
The AP will be 3, 5, 7, 9....
13. The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28th
term of this AP. [CBSE (F) 2014]
Sol. Sum of first seven terms,
Sn = n [2a + (n – 1)d]
2
S7 = 7 [2a + (7 – 1)d] = 7 [2a + 6d]
2 2
63 – 21d
63 = 7a + 21d ⇒ a= 7 ...(i)
S14 = 14 [2a + 13d] ⇒ S14 = 7 [2a + 13d] = 14a + 91d
2
But according to question, S1–7 + S8–14 = S14
116 Xam idea Mathematics–X
63 + 161 = 14a + 91d ⇒ 224 = 14a + 91d
2a + 13d = 32
2c 63 – 21d m + 13d = 32 ⇒ 126 – 42d + 91d = 224 [From equation (i)]
7 ⇒ d = 2
⇒ 49d = 98
⇒ a = 63 – 21 # 2 = 63 – 42 = 21 = 3
⇒ 7 7 7
a28 = a + 27d = 3 + 27 × 2 ⇒ a28 = 3 + 54 = 57
14. If the ratio of the sum of first n terms of two AP’s is (7n + 1) : (4n + 27), find the ratio of their
mth terms. – 1 [CBSE (AI) 2016]
n n 2
Sn 2 (2a + (n – 1) d 7n + 1 a+ d 7n + 1
Sln 4n + 27 & al + 4n + 27
Sol. = n = n – 1 = ...(i)
2 + 2
(2al (n – 1) dl) dl
Since tm = a + (m – 1) d , So replacing n –1 by m – 1 ⇒ n by 2m – 1 in (i)
tlm al + (m – 1) dl 2 ⇒
= a + (m – 1) d = 7 (2m – 1) + 1 tm = 14m – 6
al + (m – 1) dl 4 (2m – 1) + 27 tlm 8m + 23
15. Find the sum of the following series :
5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + ... + (– 5) + 81 + (– 3) [CBSE (F) 2017]
Sol. The series can be rewritten as,
(5 + 9 + 13 + ... + 81) + (–41 + (–39) + (–37) + ... + (–5) + (–3))
For the series 5 + 9 + 13 + ... 81
a = 5, d = 4 and an = 81
nth term = a + (n – 1)d = an
⇒ 5 + (n – 1)4 = 81
⇒ 4n = 80
⇒ n = 20
Sum of 20 terms for this series.
Sn = 20 (5 + 81) = 860 <a Sn = n (a + an)F …(i)
2 2
For the series (–41) + (–39) + (–37) … + (–5) + (–3)
a = –41, d = 2 and an = –3
nth term = a + (n – 1)d = an
⇒ –41 + (n – 1)2 = –3
⇒ 2n = 40
⇒ n = 20
Sum of 20 terms for this series
Sn = 20 (–41 – 3) = – 440 …(ii)
2
By adding (i) and (ii), we get
Sum of series = 860 – 440
= 420
16. Find the sum of all two digit natural numbers which are divisible by 4. [CBSE Delhi 2017 (C)]
Sol. Here a = 12, d = 4, an = 96
The formula is an = a + (n – 1)d
Therefore 96 = 12 + (n – 1) × 4
Arithmetic Progressions 117
⇒ 96 = 8 + 4n ⇒ n = 88 ⇒ n = 22
4
Apply the formula for sum,
n
Sn = 2 [2a + (n – 1) d]
Hence, S22 = 11[24 + 21 × 4] = 11[24 + 84]
= 11 × 108 = 1188.
17. In a school, students planned to plant trees in and around the school to reduce air pollution.
It was decided that number of trees, that each class will plant, will be the same as the class in
which they are studying i.e., class I will plant one tree, class II will plant two trees and so on
upto class XII.
(i) Write arithmetic progression formed by given statement.
(ii) What is first term and common difference of AP so formed?
(iii) Find total number of trees planted by school.
Sol. Since the number of trees that students plant will be same as the class in which they are studying
therefore,
(i) AP will be
1, 2, 3, . . . 12
(ii) We have, first term = 1
and common difference = 2 – 1 = 1
i.e., a = 1, d = 1
(iii) Total number of trees planted by school
sn = n (a + l)
2
= 12 (1 + 12) = 6 × 13 = 78
2
Long Answer Questions [5 marks]
1. The first term of an AP is 3, the last term is 83 and the sum of all its terms is 903. Find the number
of terms and the common difference of the AP. [CBSE 2019 (30/1/2)]
Sol. We have an AP, in which
a = 3, last term, an = l = 83
⇒ an = 83
a + (n – 1)d = 83 ⇒ 3 + (n – 1)d = 83
⇒ (n – 1)d = 83 – 3 = 80
⇒ (n – 1)d = 80 …(i)
n
Also, Sn = 903 ⇒ 2 (a + l) = 903
⇒
n (3 + 83) = 903
2
n
⇒ 2 × 86 = 903 ⇒ 43n = 903
⇒ n= 903 = 21
43
∴ n = 21
Putting n = 21 in equation (i), we get
118 Xam idea Mathematics–X
(n – 1)d = 80 ⇒ 20d = 80 ⇒ d=4
No. of terms =21 and common deference = 4
2. Divide 56 into four parts which are in AP such that the ratio of product of extremes to the
product of means is 5 : 6. [CBSE (F) 2016]
Sol. Let the four parts be a – 3d, a – d, a + d, a + 3d.
Given, (a – 3d) + (a – d) + (a + d) + (a + 3d) = 56
⇒ 4a = 56 or a = 14
Also, (a – 3d) (a + 3d) = 5
(a – d) (a + d) 6
⇒ a2 – 9d2 = 5 ⇒ 6(196 – 9d2) = 5(196 – d2) [ a = 14]
⇒ a2 – d2 6
6 × 196 – 54d2 = 5 × 196 – 5d2
⇒ 49d2 = 6 × 196 – 5 × 196 = 196
⇒ d2 = 4 or d = ±2
\ Required parts are 14 – 3×2, 14 – 2, 14+2, 14 + 3×2
or 14 – 3(–2), 14 + 2, 14 – 2, 14 + 3(–2)
i.e., 8, 12, 16, 20 or 20, 16, 12, 8
3. In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565.
Find the AP. [CBSE Delhi 2014, (F) 2017]
Sol. Let ‘a’ be the first term and ‘d’ be the common difference.
nth term of AP is an = a + (n – 1)d
n
and sum of AP is Sn = 2 [2a + (n – 1)d]
Sum of first 10 terms = 210 = 10 [2a + 9d]
2
⇒ 42 = 2a + 9d ⇒ 2a + 9d = 42 ...(i)
15th term from the last = (50 – 15 + 1)th =36th term
⇒ a36 = a + 35d 15
2
Sum of last 15 terms = 2565 = [2a36 + (15 – 1)d]
⇒ 2565 = 1=251[721(a + 35 d) + 1 4d] ⇒ 2565 = 15[a + 35d + 7d]
⇒ a + 42d
...(ii)
(i) – 2 × (ii), we get
9d – 84d = 42 – 342 ⇒ 75d = 300
⇒ tdh=e va37l0u50e =4 (ii)
Putting of d in
42 × 4 + a = 171 ⇒ a = 171 – 168
⇒ a = 3 ⇒ a50 = a + 49d = 3 + 49 × 4 = 199
So, the AP formed is 3, 7, 11, 15, ...... and 199.
4. If Sn denotes the sum of the first n terms of an AP, prove that S30 = 3 (S20 – S10). [CBSE (F) 2014]
Sol. We know that Sn = n [2a + (n – 1)d]
2
30
⇒ S30 = 2 [2a + 29d] ⇒ S30 = 30a + 435d ...(i)
Arithmetic Progressions 119
⇒ 20 S20 = 20a + 190d
S20 = 2 [2a + 19d] ⇒
⇒ S10 = 10 [2a + 9d] ⇒ S10 = 10a + 45d
2
⇒ 3(S20 – S10) = 3[20a + 190d – 10a – 45d]
⇒ = 3[10a + 145d] = 30a + 435d = S30 [From (i)]
Hence, S30 = 3(S20 – S10) Hence proved.
5. A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after
the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases
his speed by 10 m/minute every succeeding minute. After how many minutes the policeman
will catch the thief? [CBSE (Delhi) 2016]
Sol. Let total time be n minutes.
Total distance covered by thief = 100 n metres
Total distance covered by policeman = 100 + 110 + 120 + ... + (n – 1) terms
\ 100n = n –1 [100(2) + (n – 2)10]
2
⇒ 200n = (n – 1)(180 + 10n)
⇒ 10n2 – 30n – 180 = 0
⇒ n2 – 3n – 18 = 0
⇒ (n – 6) (n + 3) = 0
⇒ n = 6
Policeman took (n – 1) = (6 – 1) = 5 minutes to catch the thief.
6. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value
of X such that sum of numbers of houses preceeding the house numbered X is equal to sum of
the numbers of houses following X. Find value of X. [CBSE (AI) 2016]
Sol. The numbers of houses are 1, 2, 3, 4..........49.
The numbers of the houses are in AP, where a = 1 and d = 1
n
Sum of n terms of an AP = 2 [2a + (n – 1)d]
Let Xth number house.
house be the required
Sum of number of houses preceding Xth house is equal to SX–1 i.e.,
X –1 X –1
SX–1 = 2 [2a + (X – 1 – 1)d] ⇒ SX–1 = 2 [2 + (X – 2)]
SX–1 = X –1 (2 + X – 2) ⇒ SX–1 = X (X – 1)
2 2
Sum of numbers of houses following Xth house is equal to S49 – SX
= 49 [2a + (49 – 1)d] – X [2a + (X – 1)d]
2 2
= 49 (2 + 48) – X (2 + X – 1) = 49 (50) – X (X + 1)
2 2 2 2
= 25(49) – X (X + 1)
2
Now, we are given that
Sum of number of houses before X is equal to sum of number of houses after X.
i.e., SX–1 = S49 – SX
120 Xam idea Mathematics–X
⇒ X (X – 1) = 25(49) – X (X +1) ⇒ X2 – X = 1225 – X2 – X
2 2 2 2 2 2
⇒ X2 = 1225 ⇒ X = 1225
⇒ X = ±35
Since number of houses is positive integer, \ X = 35
7. If the ratio of the 11th term of an AP to its 18th term is 2 : 3, find the ratio of the sum of the
first five terms to the sum of its first 10 terms. [CBSE Delhi 2017 (C)]
Sol. Given, a11 = a +10d = 2 [Using formula an = a + (n – 1)d]
a18 a +17d 3
⇒ 3a + 30d = 2a + 34d ⇒ a = 4d …(i)
S5 5 (2a + 4d) n
S10 2 2
= <Using formula Sn = [2a +(n – 1) d]F
5 (2a + 9d)
= 8d + 4d (From (i))
2 (8d + 9d)
= 12d = 6
34d 17
Hence S5 : S10 = 6 : 17.
8. Students of class VIII at Navodaya Vidyalaya went on a tour to Gandey village. There were
64 students, and they were divided into equal groups to ride on 4 buses to the village. When
they go to the village all the students were divided equally into 8 groups and were allotted a
number 1 to 8. The first group started their tour at 8:00 am and each successor group started
their tour after every 10 minutes from predecessor group.
First, the students went to the Kasturba Balika Vidyalaya and visited their garden. There were
10 flower beds each containing 6 rose flowers of different colour. Garden also had fruit trees
in which half of total number of trees were of mango, One third of total were of guava and one
fourth of total were of coconut. One student named Reshma had counted coconut trees and
declared later that the number of coconut trees was 15. Next, the students visited classes and
houses of this vidyalaya. Vidyalaya had 12 classrooms for teaching purpose, 4 dormitories
for residing purpose of students and 18 staff quarters. After completion of visit all 8 groups
assemble together in a field. Scout teacher of the tour decided to give trophy to each group
such that each students of first group got one trophy, each students of 2nd group got two
trophies, each students of 3rd group got three trophies and so on. Lastly all students got bus
to return own vidyalaya.
(a) How many students ride on each bus?
(b) How many students were in each group?
(c) What time did the 5th group start their tour?
(d) How many rose flowers were in garden?
(e) How many mango trees were in garden?
(f) How many trophies were received by 8th group?
(g) How many total number of trophies were distributed among students?
Sol. (a) No. of students riding on each bus = 64 = 16 students
4
(b) No. of students in each group = 64 = 8 students
8
Arithmetic Progressions 121
(c) 8:00 am + 10 × 4 minutes = 8:40 am
(d) No. of rose flowers in garden = 10 × 6 = 60
(e) Let the no. of total trees be x .
From question,
1 of x =15
4
⇒ 1 × x =15
4
x = 15 × 4 = 60
No. of mango tree = 1 × 60 = 30
2
(f) No. of trophies received by 8th group = 8 × 8 = 64
(g) Total number of trophies distributed among students:
= 8 + 16 + 24 + 32 + 40 + 48 + 56 +64
8
= 2 {2×8 + (8 – 1) 8} Sn = n {2 × a + (n − 1)d}
2
= 4 {16 + 7 × 8}
= 4 {16 + 56}
= 4 × 72 = 288
9. In a school, on the occasion of Republic day, several athletics events are organized. In an
apple race, a bucket is placed at the starting point, which is 5 m from the first apple, and the
other apples are placed 3 m apart in a straight line. There are ten apples in the line.
During competition, Reshma starts from the bucket, picks up the nearest apple, runs back
with it, drops it in the bucket, runs back to pick up the next apple, runs to the bucket to drop
apple in it, and she continues in the same way until all the apples are in the bucket.
What is the total distance the competitor has to run?
Sol. We have,
Distance run to pick the first apple
=2×5
= 10 m (both sides going + coming)
Distance run to pick the second apple
= 2 (5 + 3)
= 16 m (both sides)
Hence, the series we have
10, 16, 22, . . . to 10 terms
Which are in AP with a = 10, d = 6
Now, the total distance the competitor has to run is given by the formula Sn = n [2a + (n – 1) d]
2
S10 = 10 [2 ×10 + (10 – 1) × 6]
2
= 5 (20 + 54) = 5 × 74 = 370 m
Total distance = 370 m.
122 Xam idea Mathematics–X
HOTS [Higher Order Thinking Skills]
1. Find the sum of the first 15 multiples of 8. [CBSE Delhi 2017 (C)]
Sol. The first 15 multiples of 8 are
8, 16, 24, ... 120
Clearly, these numbers are in AP with first term a = 8 and common difference, d = 16 – 8 = 8
Thus, S15 = 15 [2 × 8 + (15 – 1) × 8]
2
= 15 [16 + 14 × 8] = 15 [16 + 112] = 15 × 128 = 15 × 64 = 960
2 2 2
2. Find the sum of all two digit natural numbers which when divided by 3 yield 1 as remainder.
Sol. Two digit natural numbers which when divided by 3 yield 1 as remainder are:
10, 13, 16, 19, ......, 97, which forms an AP.
with a = 10, d = 3, an = 97
an = 97 ⇒ a + (n – 1) d = 97
or 10 + (n – 1)3 = 97 ⇒ (n – 1) = 87 = 29 ⇒ n = 30
3
30
Now, S30 = 2 [2 × 10 + 29 × 3] = 15(20 + 87) = 15 × 107 = 1605
3. A sum of `700 is to be used to give seven cash prizes to students of a school for their overall
academic performance. If each prize is `20 less than its preceding prize, find the value of each
of the prizes.
Sol. Let the prizes be a + 60, a + 40, a + 20, a, a – 20, a – 40, a – 60
Therefore, the sum of prizes is
a + 60 + a + 40 + a + 20 + a + a – 20 + a – 40 + a – 60 = 700
⇒ 7a = 700 ⇒ a= 700 =100
7
Thus, the value of seven prizes are
100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60
i.e., `160, `140, `120, `100, `80, `60, `40
4. If the mth term of an AP is 1 and nth term is 1 , then show that its (mn)th term is 1.
n m
[CBSE Delhi 2017, 2019(C) (30/1/1)]
Sol. Let a and d be the first term and common difference respectively of the given AP. Then
am = a + (m – 1) d ⇒ a + (m – 1)d = 1 ...(i)
n
and an = a + (n – 1) d ⇒ a + (n – 1)d = 1 ...(ii)
m
Subtracting (ii) from (i), we have
(m – n) d = 1 – 1 ⇒ (m – n) d = m–n
\ n m mn
1
d= mn
Putting d = 1 in (i), we get
mn
Arithmetic Progressions 123
a + (m – 1) 1 = 1 ⇒ a+ m – 1 = 1
mn n mn mn n
⇒ a– 1 = 0 or a= 1
mn mn
\ mnth term = a + (mn – 1) d = 1 + (mn – 1) 1 = 1+ mn – 1
mn mn mn
⇒ (mn)th term = 1
5. If the sum of m terms of an AP is the same as the sum of its n terms, show that the sum of its
(m + n) terms is zero. [CBSE Delhi 2017]
Sol. Let a be the first term and d be the common difference of the given AP.
Then, Sm = Sn n
m 2
⇒ 2 {2a + (m – 1) d} = {2a + (n – 1) d}
⇒ 2a(m – n) + {m(m – 1) – n(n – 1)}d = 0
⇒ 2a(m – n) + {(m2 – n2) – (m – n)}d = 0
⇒ (m – n) {2a + (m + n – 1) d} = 0
⇒ 2a + (m + n – 1) d = 0 [ m – n ≠ 0] ...(i)
Now, Sm + n = m+n {2a + (m + n – 1) d}
2
m+n
⇒ Sm + n = 2 × 0 = 0 [Using (i)]
6. The ratio of the sums of m and n terms of an AP is m2 : n2. Show that the ratio of the mth and
nth terms is (2m – 1) : (2n – 1).
Sol. Let a be the first term and d the common difference of the given AP. Then, the sums of m and n
terms are given by n
m 2
Sm = 2 {2a + (m – 1) d} and Sn = {2a + (n – 1) d} respectively.
Then, Sm = m2
Sn n2
⇒ m {2a + (m – 1) d} = m2 ⇒ 2a + (m – 1) d = m
2 {2a +(n – 1) d} n2 2a + (n – 1) d n
n
2
⇒ {2a + (m – 1) d} n = {2a + (n – 1) d} m
⇒ 2a(n – m) = d{(n – 1) m – (m – 1)n}
⇒ 2a(n – m) = d(n – m) ⇒ d = 2a
\ am = a +(m – 1) d = a +(m – 1) 2a = 2m –1
an a + (n – 1) d a +(n – 1) 2a 2n –1
7. If the sum of the first p terms of an AP is q and the sum of the first q terms is p; then show that
the sum of the first (p+q) terms is –(p+q). [CBSE 2019 (30/5/1)]
Sol. Let first term be a and common difference of the AP be d
p
Given, Sp = q ⇒ 2 (2a + (p – 1)d) = q
⇒ 2a + (p – 1)d = 2q ...(i)
p q
2 {2a + (q – 1)d} = p
Also Sq = p ⇒
124 Xam idea Mathematics–X
2p ...(ii)
⇒ 2a + (q – 1)d = q
On subtracting (ii) from (i), we have
(p – q)d = 2q − 2p = 2(q2 − p2)
p q pq
⇒ –2 (p + q) ...(iii)
d = pq
On putting the value of d in (i), we get
2a + (p – 1) × * –2_ p+ qi 4 = 2q
pq p
⇒ a= q + _p + qi^ p–1h ...(iv)
p pq (from (iii) and (iv))
p+q
Now, Sp+q = 2 (2a + (p + q – 1)d)
= p + q f2 q + 2_ p + qi^ p–1h + (p + q – 1) × –2 (p + q) p
2 p pq pq
= _p + qif q + _ p + qi^ p–1h – (p + q – 1)_ p + qi
p pq pq p
= _p + qif q _p + qi = – (p+q)
p – p p
\ Sp+q = – (p + q)
8. If a2, b2, c2, are in AP, prove that b a c , c b a , c are in AP.
Sol. Since a2, b2, c2, are in AP. + + a+b
\ b2 – a2 = c2 – b2 ...(i)
Now, b – a = b2 + bc – ac – a2
c+a b+c (a + c) (b + c)
= (b2 – a2)+ c (b – a) = (b – a) (b + a + c) ...(ii)
(a+ c) (b+ c) (a + c) (b + c)
Also, c – b a = c2 + ac – ab – b2 = c2 – b2 + a (c – b) = (c – b) (c + b + a)
a+b c+ (a+ b) (c+ a) (a + b) (c + a) (a + b) (a + c)
= (c2 – b2) (c + b + a)
(a + b) (b + c) (c+ a)
[Multiplying numerator and denominator by b + c ]
= (b2 – a2) (a + b + c) [Using (i)]
(a + b) (b + c) (c+ a)
= (b – a) (a + b + c) ...(iii)
From (ii) and (iii), we have (b+ c) (c+ a)
b – a = c – b
c+a b+c a+c c+a
i.e., b a c , c b a , c are in AP.
+ + a+b
Arithmetic Progressions 125
PROFICIENCY EXERCISE
QQ Objective Type Questions: [1 mark each]
1. Choose and write the correct option in each of the following questions.
(i) Sum of first five multiples of 5 is
(a) 45 (b) 55 (c) 65 (d) 75
(ii) Which term of the AP 21, 42, 63, 84 _______ is 210?
(a) 9th (b) 10th (c) 11th (d) 12th
(iii) The nth term of an AP is given by an = 3 + 4n. The common difference is
(a) 7 (b) 3 (c) 4 (d) 1
(iv) If p, q, r and s are in AP, then r – q is
(a) s – p (b) s – q (c) s – r (d) None of these
(v) The sum of 12 terms of an AP, whose nth term is given by an = 3n + 4 is
(a) 262 (b) 272 (c) 282 (d) 292
2. Fill in the blanks.
(i) The sum of first ‘n’ odd natural numbers is ___________ .
(ii) If a, b, c are in AP then a –b is equal to ___________ .
b–c
(iii) nth term of the sequence a, a + d, a + 2d... is ___________.
(iv) Fill the two blanks in the sequence 3, _______ 9 _______ so that the sequence forms an AP.
(v) 21st term of an AP with first term p and common difference q is ______________.
QQ Very Short Answer Questions: [1 mark each]
3. Find the sum of first 10 multiples of 6. [CBSE 2019 (30/3/1)]
4. Find the common difference of the AP
1 , 3– a , 3 – 2a , . . . ^a ! 0h [CBSE 2019 (30/2/1)]
a 3a 3a
5. How many two digits numbers are divisible by 3? [CBSE 2019 (30/1/1)]
6. Find the number of terms in the AP 18, 15 1 , 13, ..., – 47. [CBSE 2019 (30/4/2)]
2
7. Which term of the AP – 4, – 1, 2, ... is 101? [CBSE 2019 (30/4/3)]
8. If in an AP, a = 15, d = – 3 and an = 0, then find the value of n. [CBSE 2019 (30/5/1)]
9. If the nth term of an AP is pn + q, find its common difference. [CBSE 2019 (C) (30/1/1)]
10. Which term of the AP 4, 7, 10, ... is 64? [CBSE 2019 (C) (30/1/2)]
11. Which term of the AP 10, 7, 4, ... is – 41? [CBSE 2019 (C) (30/1/3)]
12. The 9th term of an AP is 449 and 449th term is 9. Find the term which is equal to zero.
13. Find the common difference if nth term of AP is 7 – 4n.
14. If the sum of first n even natural numbers is equal to k times the sum of first n odd natural
numbers, then find k.
15. How many terms of the AP 27, 24, 21 ....... should be taken so that their sum is zero?
[CBSE Delhi 2016]
126 Xam idea Mathematics–X
16. For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k – 1 form an AP?
[CBSE (F) 2016]
1 1 + m 1 + 2m
17. Write the nth term of the AP m , m , m , ......... [CBSE Delhi 2017 (C)]
QQ Short Answer Questions–I: [2 marks each]
18. If Sn, the sum of first n terms of an AP is given by Sn = 3n2 – 4n, find the nth term.
[CBSE 2019 (30/1/1)]
19. How many multiples of 4 lie between 10 and 205? [CBSE 2019 (30/4/2)]
20. Determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
[CBSE 2019 (30/4/2)]
21. Find the sum of 7 + 10 + 13 + ... + 46. [CBSE 2019 (C) (30/1/1)]
22. If the 9th term of an AP is zero, then show that its 29th term is double of its 19th term.
[CBSE 2019 (C) (30/1/1)]
23. For what values of ‘k’, k – 3, 2k + 1 and 4k + 3 are in AP?
24. Determine k so that k2 + 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP.
25. Write the value of a25 – a15 for the AP 5, 9, 13, 17, .................
26. Write 7th term from the end of the AP 7, 9, 11, 13, ........... 213.
27. The general term of a sequence is given by an = –4n + 15. Is the sequence forming an AP? If so,
find its 15th term and the common difference.
28. Find whether –150 is a term of the AP 11, 8, 5, 2, …….. [CBSE Delhi 2017 (C)]
29. How many terms are there in the AP –1, –5 , –2 , –1 , ........... 10 ?
6 3 2 3
30. In an AP, prove that am+n + am–n = 2am where an denotes its nth term.
Which of the following form an AP? Justify your answer (Q. 31 to 32)
31. 0, 0, 1, 1, 2, 2, 3, 3 ..............
32. 2, 22, 23, 24.............
33. Is 0 a term of the AP 31, 28, 25,...........? Justify your answer.
34. Is the sum of m terms of an AP always less than the sum of (m + 1) terms? Give reason.
35. For the AP –3, –7, –11...... can we find directly a30 – a20 without actually finding a30 and a20? Give
reason for your answer.
36. The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420, find
its common difference. [CBSE Delhi 2014]
37. The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th
term, find the AP. [CBSE (AI) 2014]
38. The sum of the first n terms of an AP is 4n2 + 2n. Find the nth term of this AP. [CBSE (F) 2014]
39. The 10th term of an AP is (– 4) and its 22nd term is (–16). Find its 38th term. [CBSE Delhi 2017 (C)]
QQ Short Answer Questions–II: [3 marks each]
40. In an AP if sum of its first n terms is 3n2 + 5n and its kth term is 164, find the value of k.
5 1
41. Which term of the AP, 6 , 1, 1 6 ..... is 3?
42. An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the
last three is 429. Find the AP.
Arithmetic Progressions 127
43. Which term of the AP, 8, 14, 20, 26, ...... will be 72 more than its 41st term? [CBSE (AI) 2017]
44. Determine the AP whose 5th term is 19 and the difference of the 8th term and the 13th
term is 20. [NCERT Exemplar]
45. How many multiples of 5 lie between 50 and 250?
46. How many three digit numbers are divisible by 9?
47. Which term of the AP 3, 12, 21, 30,…. will be 90 more than its 50th term? [CBSE Delhi 2017 (C)]
48. If 9th term of an AP is zero, prove that its 29th term is double the 19th term. [NCERT Exemplar]
49. The sum of the 6th and 9th terms of an AP is 101 and the sum of the 10th and 16th terms is 178.
Find the first three terms of the AP.
50. If the sum of the first n terms of an AP is 4n – n2, what is the 10th term and the nth term?
51. Find the sum of all three digit natural numbers, divisible by 7.
52. Solve the equation: – 4 + (– 1) + 2 +....+ x = 437 [NCERT Exemplar]
53. If 12th term of an AP is –13 and the sum of the first four terms is 24, what is the sum of first 10
terms?
54. Yasmeen saves `32 during the first month, `36 in the second month and `40 in the third month.
If she continues to save in this manner, in how many months will she save `2000?
[NCERT Exemplar]
55. Find the sum:
(i) a4 – 1 k + a4 – 2 k + a4 – 3 k + .... upto n terms. [CBSE Delhi 2017]
n n n
(ii) a–b + 3a – 2b + 5a – 3b + ........... to 20 terms.
a+b a+b a+b
56. Which term of the AP –2, –7, –12,...will be –77? Find the sum of this AP upto the term –77.
[NCERT Exemplar]
57. Split 207 into three parts such that these are in AP and the product of the two smaller parts is
4623. [NCERT Exemplar]
QQ Long Answer Questions: [5 marks each]
58. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and
the last term to the product of two middle terms is 7 : 15. Find the numbers. [CBSE Delhi 2018]
59. If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its
first n terms. [CBSE 2019 (30/1/3)]
60. Which term of the Arithmetic Progression – 7, – 12, – 17, – 22, ... will be – 82? Is – 100 any term
of the AP? Give reason for your answer. [CBSE 2019 (30/2/1)]
61. How many terms of the Arithmetic Progression 45, 39, 33, ... must be taken so that their sum is
180? Explain the double answer. [CBSE 2019 (30/2/1)]
62. If m times the mth term of an Arithmetic Progression is equal to n times its nth term show that the
(m + n)th term of the AP is zero. [CBSE 2019 (30/3/1)]
63. The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first
and the third term is 5 times the common difference, find the three numbers.
[CBSE 2019 (30/3/1)]
64. Find the value of x, when in the AP given below 2 + 6 + 10 + ... + x = 1800.
[CBSE 2019 (30/4/2)]
128 Xam idea Mathematics–X
65. In an AP, the first term is – 4, the last term is 29 and the sum of all its terms is 150. Find its
common difference. [CBSE 2019 (30/4/3)]
66. Find the sum of all the two digit numbers which leave the remainder 2 when divided by 5.
[CBSE 2019 (30/5/2)]
67. Find the sum of all odd numbers between 0 and 50. [CBSE 2019 (C) (30/1/1)]
68. Find the sum of the integers between 100 and 200 that are: (i) divisible by 9 (ii) not divisible
by 9. [NCERT Exemplar]
69. Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal
^a + ch^b +c– 2ah
to 2^b – ah . [NCERT Exemplar]
70. Sums of the first p, q, r terms of an AP are a, b, c respectively. Prove that
a (q − r) + b (r − p) + c (p − q) = 0
p q r
71. If the pth term of an AP is 1 and qth term is 1 , prove that the sum of the pq terms is 1 (pq + 1).
q p 2
[CBSE Delhi 2017]
72. The sum of the first five term of an AP is 55 and sum of the first ten terms of this AP is 235, find
the sum of its first 20 terms.
73. The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the
21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
[NCERT Exemplar]
74. Three hundred sixty bricks are stacked in the following manner: 30 bricks in the bottom row, 29
in the next row, 28 in the row next to it and so on. In how many rows are the 360 bricks placed
and how many bricks are there in the top row?
75. The students of a school decided to beautify the school on the Annual Day by fixing colourful flags
on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The
flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing
the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time.
How much distance did she cover in completing this job and returning back to collect her books?
What is the maximum distance she travelled carrying a flag? [NCERT Exemplar]
76. A contract on construction job specifies a penalty for delay of completion beyond a certain days
as follows: ™180 for the first day, ™200 for the second day, etc., the penalty for each succeeding
day being ™20 more than the preceding day.
(i) How much money the contractor has to pay as penalty, if he has delayed the work by 15 days?
(ii) For how many days has the contractor delayed the work if he paid ™ 7400 as penalty?
77. Find the middle term of the sequence formed by all three-digit numbers which leave a remainder
3, when divided by 4. Also find the sum of all numbers on both sides of the middle term
separately. [CBSE (F) 2015]
78. Find the sum of all the two digit numbers which are either multiples of 2 or 3.
79. Ramkali required ™2500 after 12 weeks to send her daughter to school. She saved ™100 in the
first week and increased her weekly saving by ™20 every week. Find whether she will be able to
send her daughter to school after 12 weeks. [CBSE Delhi 2015]
80. Find the middle term of the sequence formed by all numbers between 9 and 95, which leave a
remainder 1 when divided by 3. Also find the sum of the numbers on both sides of the middle
term separately. [CBSE (F) 2015]
Arithmetic Progressions 129
81. The pth, qth and rth terms of an AP are a, b and c respectively. Show that a(q – r) + b(r – p) +
c(p – q) = 0. [CBSE (F) 2016]
82. If the ratio of the sum of the first n terms of two APs is (7n + 1) : (4n + 27), then find the ratio of
their 9th terms.
[CBSE (AI) 2017]
Answers
1. (i) (d) (ii) (b) (iii) (c) (iv) (c) (v) (c)
2. (i) n2 (ii) 1 (iii) an = a + (n – 1) d (iv) 6 ; 12 (v) p + 20q
1 n = 30 n = 36 8. n = 6
3. 330 4. – 3 5. 6. n = 27 7.
19. n = 49
9. d = p 10. n = 21 11. n = 18 12. 458th term 13. d = – 4 25. 40
n +1 1 31. No
14. k = n 15. n = 19 16. k = 6 17. m + n – 1 18. 6 n – 7
20. AP is 4, 10, 16 21. Sn = 371 23. k = 2 24. k = 0
26. 201 27. Yes, – 45; – 4 28. No 29. 27
32. No 33. No 34. No 35. Yes, we can find it using ‘d’ only
36. 3 37. 1, 5, 9.... 38. 8n – 2 39. a38 = – 32 40. k = 27 41. 14th term
43. 53rd term 44. 3, 7, 11, 15 .... 45. 39
42. 3, 7, 11,15.....
46. n = 100 47. 60th term 49. 5, 12, 19 50. –15, 5 – 2n 51. 70336 52. x = 50
53. 0
54. 25 months 55. (i) 7n – 1 (ii) 10 (40a – 21b)
2 a+b
56. 16th term, – 632 57. 67, 69, 71 58. Numbers 2, 6, 10, 14 or 14, 10, 6, 2
59. n(n + 6) or (n2 + 6n) 60. n = 16, – 100 is not a term of AP
61. n = 6 or n = 10 63. 15, 6, – 3 or 2, 6, 10 64. x = 118 65. d = 3
66. Sn = 981 67. 625 68. (i) 1683 (ii) 13167 72. 970 73. 1 : 3; 5 : 49
74. 16 rows; 15 bricks 75. 728 m, 26 m
76. (i) ™4800 (ii) 20 days
77. 551, 36400, 87024 78. 3285 79. 2520; yes 80. 52; 413, 1043 82. 24 : 19
SELF-ASSESSMENT TEST
Time allowed: 1 hour Max. marks: 40
Section A
1. Choose and write the correct option in the following questions. (4 × 1 = 4)
(i) The number of terms in the AP 2, 5, 8, ..., 59 is
(a) 12 (b) 19 (c) 20 (d) 25
(ii) The sum of first 20 odd natural numbers is
(a) 281 (b) 285 (c) 400 (d) 421
(iii) The 4th term from the end of the AP –11, – 8, – 5, ... 49 is
(a) 37 (b) 40 (c) 43 (d) 58
(iv) If the first term of an AP is –5 and the common difference is 2, then the sum of the first
6 terms is [NCERT Exemplar]
(a) 0 (b) 5 (c) 6 (d) 15
130 Xam idea Mathematics–X
2. Fill in the blanks. (3 × 1 = 3)
(i) The nth term of an AP is always a ____________ expression.
(ii) The 12th term of an AP with first term a and common difference d is ____________.
(iii) The 6th term from the end of the AP 17, 14, 11, ........... –40 is ____________.
3. Solve the following questions. (3 × 1 = 3)
(i) What is the nth term if the sum of n term of an AP is n2 – n?
(ii) How many two digit numbers are divisible by 3?
(iii) Which is the first positive term of the AP –11, –8, –5...?
Section B
QQ Solve the following questions. (3 × 2 = 6)
4. If the 10th term of an AP is 52 and 17th term is 20 more than the 13th term, find the AP.
5. If the pth term of an AP is q and the qth term is p, prove that its nth term is (p + q – n).
6. Find a, b and c such that the numbers a, 7, b, 23, c are in AP. [NCERT Exemplar]
QQ Solve the following questions. (3 × 3 = 9)
7. In an AP prove that am+n + am–n = 2am, where an denotes its nth term. [CBSE (F) 2017]
8. If 1 + 4 + 7 + 10 + ... + x = 287, find the value of x.
9. Jaspal singh repays his total loan of ™118000 by paying every month starting with the first
instalment of ™1000. If he increases the instalment by ™100 every month, what amount will be
paid by him in the 30th instalment? What amount of loan does he still have to pay after the 30th
instalment? [NCERT Exemplar]
QQ Solve the following questions. (3 × 5 = 15)
10. I f mth ter m of an AP is 1n and nt h term is m1 , then find th e sum of its first mn terms.
[CBSE Delhi 2017]
11. An arithmetic progression 5, 12, 19, ..., has 50 terms. Find its last term. Hence find the sum of
its last 15 terms. [CBSE (AI) 2015]
12. The sum of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The
first term of each AP is 1 and their common differences are 1, 2 and 3 respectively. Prove that
S1 + S3 = 2S2. [CBSE (AI) 2016]
Answers
1. (i) (c) (ii) (c) (iii) (b) (iv) (a)
2. (i) linear (ii) a+11d (iii) –25 (iv) 1
3. (i) 2n – 2 (ii) 30 (iii) 1
4. 7, 12, 17, 22.... 6. a = –1, b = 15, c = 31 8. x = 40
9. ™3,900; ™44,500 11. 348; 4485
10. 1 (mn+1)
2
zzz
Arithmetic Progressions 131
6 Coordinate
Geometry
BASIC CONCEPTS – A FLOW CHART
132 Xam idea Mathematics–X
MORE POINTS TO REMEMBER
For a given point, the abscissa and ordinate are the distance of the given point from y-axis
and x-axis respectively.
For example, the point P(3, 2) is at distance 3 unit from
y-axis and 2 unit from x-axis.
The x co-ordinate or abscissa of every point on y-axis is
zero, while y-coordinate or ordinate of every point on
x-axis is zero. The coordinate of origin is (0, 0).
Fig. 6.1
Fig. 6.2
Use of distance formula:
Collinear points: To show collinearity of three points, we prove that the sum of the
distance between two pair of points is equal to the third pair of points. For example, points
A, B, C are collinear because AB + BC = AC.
Fig. 6.3
Equidistance: To prove that a point ‘P’ is equidistance from two
points A and B, we prove PA = PB.
Geometrical figure: If three or four points are given, then in order
to prove that a given figure is:
(i) triangle, we prove that sum of lengths of any two sides is greater Fig. 6.4
than the length of third side.
(ii) isosceles triangle, we prove that any two sides are equal.
(iii) equilateral triangle, we prove that all three sides are equal.
(iv) right angled triangle, we prove that sum of square of two sides are equal to square of
the third longest side.
(v) square, we prove that four sides are equal and the diagonals are also equal.
(vi) rhombus, we prove that four sides are equal.
(vii) rectangle, we prove that opposite sides are equal and the diagonals are also equal.
(viii) parallelogram, we prove that opposite sides are equal.
(ix) parallelogram but not a rectangle, we prove that its opposite sides are equal but
diagonals are not equal.
(x) rhombus but not a square, we prove that its all sides are equal but diagonals are not
equal.
Coordinate Geometry 133
Facts related to section formula and its use:
(i) In case of internal division, point dividing the line segment lie on it, while in external
division, point dividing the line segment does not lie on it.
For example,
Fig. 6.5
(ii) If a point P divides a line segment AB in the ratio m:n, externally then the coordinate
of point is mx2 − nx1 , my2 − ny1 , where coordinate of A and B are (x1, y1), (x2, y2)
m − n m − n
respectively.
(iii) To show the collinearity of three points A, B, C by using section formula, first we assure
that any one point say C divides AB in the ratio of k : 1. Then we find coordinate of C
by using section formula and equate them with x and y coordinate of C. If value of k
from both equation are same, then given points, A, B, C are collinear.
Collinearity of points by using area formula: To prove the collinearity of three points A,
B, C by using area formula, we find area of triangle ABC formed assuming vertices as given
points A, B and C. If area so obtained is zero, then given points are collinear.
Multiple Choice Questions [1 mark]
Choose and write the correct option in the following questions.
1. If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is
(b) 2 (c) –1 [NCERT Exemplar]
(a) –2 (d) 1
2. The mid-point of the line segment joining the points A (–2, 8) and B (–6, –4) is
(b) (2, 6) (c) (–4, 2) [NCERT Exemplar]
(a) (–4, –6) (d) (4, 2)
3. The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a [NCERT Exemplar]
(a) square (b) rectangle (c) rhombus (d) trapezium
4. The distance of the point P (2, 3) from the x-axis is [NCERT Exemplar]
(d) 5 units
(a) 2 units (b) 3 units (c) 1 units
5. The distance between the points A (0, 6) and B (0, –2) is [NCERT Exemplar]
(d) 2 units
(a) 6 units (b) 8 units (c) 4 units
6. The distance of the point P (–6, 8) from the origin is [NCERT Exemplar]
(d) 6 units
(a) 8 units (b) 2 7 units (c) 10 units
7. The distance between the points A (0, 7) and B (0, –3) is
(a) 4 units (b) 10 units (c) 7 units (d) 3 units
8. The point A (–5, 6) is at a distance of (b) 11 units from origin
(a) 61 units from origin
(c) 61 units from origin (d) 11 units from origin
134 Xam idea Mathematics–X
9. The end points of diameter of circle are (2, 4) and (–3, –1). The radius of the circle is
(a) 52 units (b) 5 2 units (c) 3 2 units (d) !5 2 units
2 2
10. The coordinates of the point which is equidistant from the three vertices of the triangle shown
in the given Fig. 6.6 are
Y
(0, 2yA)
X' O (2xB, 0) X
Y'
Fig. 6.6
(a) (x, y) (b) (y, x) (c) e x , y o (d) e y x o
2 2 2, 2
11. The coordinates x y
(a) (a, 0) of the point where line a + b = 7 intersects y-axis are
(0, 7b) (d) (7a,
(b) (0, b) (c) 0)
12. The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is [NCERT Exemplar]
(a) 14 sq. units (b) 28 sq. units (c) 8 sq. units (d) 6 sq. units
13. The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2
internally lies in the [NCERT Exemplar]
(a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant
14. The points which lie on the perpendicular bisector of the line segment joining the points
A (–2, –5), B (2, 5) is [NCERT Exemplar]
(a) (0, 0) (b) (0, 2) (c) (2, 0) (d) (–2, 0)
15. If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
(a) AP = 1 AB (b) AP = PB (c) PB = 1 AB (d) AP = 1 AB
3 3 2
16. If Pc a , 4 m is the mid-point of the line segment joining the points Q (–6, 5) and R (–2, 3), then
3
the value of a is [NCERT Exemplar]
(a) –4 (b) –12 (c) 12 (d) –6
17. The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the
y-axis at [NCERT Exemplar]
(a) (0, 13) (b) (0, –13) (c) (0, 12) (d) (13, 0)
18. A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, –5) is the mid-
point of PQ, then the coordinates of P and Q are, respectively [NCERT Exemplar]
(a) (0, –5) and (2, 0) (b) (0, 10) and (–4, 0)
(c) (0, 4) and (–10, 0) (d) (0, –10) and (4, 0)
19. The area a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is [NCERT Exemplar]
(a) (a + b + c)2 (b) 0 (c) a + b + c (d) abc
Coordinate Geometry 135
20. The distance between the lines 2x + 4 = 0 and x – 5 = 0, is
(a) 9 units (b) 1 unit (c) 5 units (d) 7 units
21. The perimeter of triangle formed by the points (0, 0), (2, 0) and (0, 2) is
(a) 4 units (b) 6 units (c) 6 2 units (d) 4 + 2 2 units
22. The distance of the point P (5, –1) from the y-axis is
(a) 5 units (b) 2 units (c) 3 units (d) 7 units
Answers 2. (c) 3. (b) 4. (b) 5. (b) 6. (c)
8. (a) 9. (a) 10. (a) 11. (c) 12. (c)
1. (b) 14. (a) 15. (d) 16. (b) 17. (a) 18. (d)
7. (b) 20. (d) 21. (d) 22. (a)
13. (d)
19. (b)
Fill in the Blanks [1 mark]
Complete the following statements with appropriate word(s) in the blank space(s).
1. The distance of a point from the x-axis is called its _______________.
2. The distance of a point from the y-axis is called its _______________.
3. A point of the form (0, a) lies on _______________.
4. A point of the form (b, 0) lies on _______________.
5. If area of the triangle formed by the vertices A (x1, y1) B (x2, y2) and C (x3, y3) is zero, then the
points A, B and C are _______________.
6. The value of the expression x2 + y2 is the distance of the point P (x, y) from the _______________.
7. All the points equidistant from two given points A and B lie on the _______________ of the line
segment AB.
8. If the co-ordinates of the points A, B, C and D are such that AB = BC = CD = DA and AC = BD,
then quadrilateral ABCD is a _______________.
9. If the coordinates of the points P, Q, R and S are such that PQ = QR = RS = SP and PR≠QS,
then quadrilateral PQRS is a _______________.
10. If the coordinates of the points D, E, F and G are such that DE = FG, EF = GD and DF = EG,
then quadrilateral DEFG is a _______________.
11. The distance of the point (p, q) from (a, b) is ____________ .
12. The fourth vertex D of a parallelogram ABCD whose three vertices are A (– 2, 5), B (6, 9) and
C (8, 5) is _____________.
13. The point which divides the line segment joining the points (5, 4) and (– 6, –7) in the ratio 1 : 3
internally lies in the _____________ quadrant.
14. The abscissa of a point in the third quadrant is always ____________. (positive/negative)
15. The centroid of a triangle divides each median in the ratio ____________.
Answers
1. ordinate 2. abscissa 3. y-axis 4. x-axis 5. collinear 6. origin
7. perpendicular bisector 8. square
9. rhombus 10. rectangle
11. (a–p)2 + (b–q)2 12. (0,1) 13. first 14. negative 15. 2 : 1
136 Xam idea Mathematics–X
Very Short Answer Questions [1 mark]
1. What is the area of the triangle formed by the points O (0, 0), A (–3, 0) and B (5, 0) ? What can
you say about collinearity?
1
Sol. Area of ∆OAB = 2 [0(0 – 1) – 3(0 – 0) + 5(0 – 0)] = 0
⇒ Given points are collinear.
2. If the centroid of triangle formed by points P (a, b), Q (b, c) and R (c, a) is at the origin, what
is the value of a + b + c ? b c, b + c+a
a 3 3
Sol. Centroid of ∆PQR = + +
Given a + b + c , b + c+ a = (0, 0)
3 3
⇒ a + b + c = 0
3. AOBC is a rectangle whose three vertices are A (0, 3), O (0, 0) and B (5, 0). Find the length of
its diagonal. [NCERT Exemplar]
Sol. Length of diagonal = AB = (5 − 0)2 + (0 − 3)2 = 25 + 9 = 34
4. Find the value of a, so that the point (3, a) lie on the line 2x – 3y = 5. [CBSE Delhi 2009]
[NCERT Exemplar]
Sol. Since (3, a) lies on the line 2x – 3y = 5
\ 2(3) – 3(a) = 5 ⇒ – 3a = 5 – 6
⇒ – 3a = – 1 1
⇒ a= 3
5. Find distance between the points (0, 5) and (– 5, 0).
Sol. Here x1 = 0, y1 = 5, x2 = –5 and y2 = 0.
\ d = (x2 − x1)2 + ( y2 − y1)2
= (–5 – 0)2 +(0 – 5)2
= 25 + 25 = 50 = 5 2 units
6. Find the coordinates of a point A, where AB is a diameter of the circle with centre (– 2, 2) and
B is the point with coordinates (3, 4). [CBSE 2019 (30/1/2)]
OR
Find the coordinaes of a point A, where AB is a diameter of the circle with centre (3, –1) and
the point B is (2, 6). [CBSE 2019 (30/5/2)]
Sol. Let co-ordinate of point A be (x, y).
x+3
∴ –2= 2 ⇒ x+3=–4 ⇒x=–7
⇒y=0
and 2= y+4 ⇒ y + 4 = 4
2
∴ Co-ordinate of A are (–7, 0).
OR
Similar solution as above. Ans.: (4, – 8)
7. If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k?
[CBSE Delhi 2017]
Sol. Using distance formula (x2 – x1)2 + (y2 – y1)2 , we have
Coordinate Geometry 137
⇒ (4 – 1)2 + (k – 0)2 = 5
⇒ 9 + k2 = 25 ⇒ k2 = 16
⇒ k = ! 4
8. If the points A (1, 2), B (0, 0) and C (a, b) are collinear, then what is the relation between a and b?
Sol. Points A, B and C are collinear.
⇒ 1(0 – b) + 0 (b – 2) + a(2 – 0) = 0
⇒ –b + 2a = 0 or 2a = b
9. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by
(– 1, 6). [NCERT]
Sol. In Fig. 6.7, let the point P(–1, 6) divides the line joining A(–3, 10) and B (6, –8) in the ratio k : 1
then, the coordinates of P are 6k − 3 , −8k + 10 .
k +1 k +1
But, the coordinates of P are (– 1, 6).
k1
\ 6k – 3 = – 1 Fig. 6.7
k+1 ⇒ 6k – 3 = – k – 1
⇒ 6k + k = 3 – 1 ⇒ 7k = 2
⇒ k= 2
7
Hence, the point P divides AB in the ratio 2 : 7.
10. The coordinates of the points P and Q are respectively (4, –3) and (–1, 7). Find the abscissa of
a point R on the line segment PQ such that PR = 3 .
PQ 5
Sol.
Fig. 6.8
PQ = 5 ⇒ PR+ RQ = 5 & 1 + RQ = 5 ⇒ RQ = 2
PR 3 PR 3 PR 3 PR 3
i.e., R divides PQ in the ratio 3 : 2.
Abscissa of R = 3 # (–1)+ 2 # 4 = –3 + 8 =1
3+2 5
Short Answer Questions-I [2 marks]
1. Write the coordinates of a point on x-axis which is equidistant from the points (–3, 4) and (2, 5).
Sol. Let the required point be (x, 0).
Since, (x, 0) is equidistant from the points (–3, 4) and (2, 5).
\ (−3 − x)2 + (4 − 0)2 = (2 − x)2 + (5 − 0)2
⇒ 9 + x2 + 6x + 16 = 4 + x2 − 4x + 25
⇒ x2 + 6x + 25 = x2 – 4x + 29
⇒ 10x = 4 or x = 4 = 2
10 5
138 Xam idea Mathematics–X
\ Required point is 2 , 0 .
5
2. In Fig. 6.9, if A (–1, 3), B(1, –1) and C (5, 1) are the vertices of a triangle ABC, what is the length
of the median through vertex A? 1 + 5, −1 + 1 = (3, 0)
Sol. Coordinates of the mid-point of BC = 2 2
\ Length of the median through A = (3 + 1)2 + (0 − 3)2
= 16 9 25 = 5 units Fig. 6.9
3. Find the ratio in which the line segment joining the points P (3, –6) and Q (5, 3) is divided by
the x-axis.
Sol. Let the required ratio be l : 1
Then, the point of division is 5λ +3 , 3λ −6 .
λ +1 λ +1
Given that this point lies on the x-axis.
\ 3λ − 6 = 0 or 3l = 6 or l = 2
λ +1
Thus, the required ratio is 2 : 1.
4. The mid point of the line segment joining A (2a, 4) and B(–2, 3b) is (1, 2a+1). Find the values
of a and b. [CBSE 2019 (30/5/1)]
Sol. As (1, 2a + 1) is the mid point of AB.
∴ 1= 2a – 2 and 2a +1= 4 + 3b
2 2
Now, 1= 2a – 2 ⇒ 2 = 2a – 2 ⇒ 2a = 4 ⇒ a = 2
2
and 2a + 1= 4 + 3b ⇒2×2+1= 4 + 3b
2 2
⇒ 10 = 4 + 3b ⇒ 3b = 10 – 4 = 6 ⇒ b = 2
∴ a = 2, b = 2
5. Show that ΔABC, where A(–2, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 4) are
similar triangles. [CBSE Delhi 2017]
OR
Show that ∆ABC with vertices A(– 2, 0), B(0, 2) and C(2, 0) is similar to ∆DEF with vertices
D(– 4, 0), F(4, 0) and E(0, 4). [NCERT Exemplar, CBSE (F) 2017]
[∆PQR is replaced by ∆DEF] AP
Sol. AB = (2 + 2)2 + 0 = 16 = 4
BC = (0 – 2)2 + (2 – 0)2 = 8 = 2 2
CA = (–2 –0)2 + (0 – 2)2 = 8 = 2 2 B CQ R
PQ = (4 + 4)2 + 0 = 64 = 8
Fig. 6.10
QR = (0 – 4)2 + (4 – 0)2 = 32 = 4 2
RP = (– 4 – 0)2 + (0 – 4)2 = 32 = 4 2
\ AB = BC = CA = 1 ⇒ ∆ABC ∼ ∆PQR
PQ QR RP 2
Coordinate Geometry 139
6. Is point P (0, 2) the point of intersection of y-axis and perpendicular bisector of line segment
joining the points, A (–1, 1) and B (3, 3) ?
Sol. The point P (0, 2) lies on y-axis.
Also, AP = (0 + 1)2 + (2 − 1)2 = 2
BP = (0 − 3)2 + (2 − 3)2 = 9 + 1 = 10
a AP ≠ BP
Hence, P(0, 2) does not lie on the perpendicular bisector of AB.
7. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. [NCERT]
Sol. Let A (5, – 2), B (6, 4) and C (7, – 2) be the vertices of a triangle.
Then we have,
AB = (6 − 5)2 + (4 + 2)2 = 1 + 36 = 37
BC = (7 − 6)2 + (−2 − 4)2 = 1 + 36 = 37
AC = (7 – 5)2 +(–2+ 2)2 = 4 = 2
Here, AB = BC Fig. 6.11
\ ∆ABC is an isosceles triangle.
8. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
[NCERT]
Sol. Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) be the vertices of a parallelogram ABCD.
Since, the diagonals of a parallelogram bisect each other.
\ x + 1 , 6 + 2 = 3 + 4 , 5+ y
⇒ 2 2 2 2
x +1 = 7
2 2
⇒ x + 1 = 7 or x = 6
⇒ 4= 5+ y or 5 + y = 8 or y=8–5=3
2
Fig. 6.12
Hence, x = 6 and y = 3.
9. Find the ratio in which y-axis divides the line segment joining the points A(5, –6) and
B(–1, –4). Also find the coordinates of the point of division. [CBSE Delhi 2016]
Sol. Let the point on y-axis be P(0, y) and AP : PB = k : 1
Therefore 5 − k = 0 gives k = 5
k + 1
Hence required ratio is 5 : 1.
y= 4(5) 6 26 13
5 1 6 3
13
Hence, point on y-axis is P § 0, 3 · . Fig. 6.13
©¨ ¸¹
10. Let P and Q be the points of trisection of the line segment joining the points A(2, – 2) and
B(–7, 4) such that P is nearer to A. Find the coordinates of P and Q. [CBSE (AI) 2016]
Sol. a P divides AB in the ratio 1 : 2.
\ Coordinates of § 1 u ( 7) 2 u 2 · , §1 u 4 2 u ( 2) ·
P = ¨ 1 2 ¸ ¨ ¸
¹ © 1 2 ¹
©
−7 + 4 4 − 4 Fig. 6.14
= 3 , 3 = (−1, 0)
a Q is the mid-point of PB.
140 Xam idea Mathematics–X
\ Coordinates of Q = e –1+ (–7) , 0 + 4 o
2 2
= c –8 , 2 m = ^– 4, 2h
2
11. Find the ratio in which the point (–3, k) divides the line-segment joining the points (–5, –4) and
(–2, 3). Also find the value of k. [CBSE (F) 2016]
Sol. Let Q divide AB in the ratio of p : 1.
–3 = −2 p − 5
p +1
⇒ –3p – 3 = –2p – 5 ⇒ p=2 Fig. 6.15
\ Ratio is 2:1.
k = 2× 3− 4 = 2
2+1 3
12. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, –5) and
R(–3, 6), find the coordinates of P. [CBSE Delhi 2016]
Sol. Let the point P be (2y, y).
PQ = PR ⇒ (2y − 2)2 + ( y + 5)2 = (2y + 3)2 + ( y − 6)2
⇒ 4y2 + 4 – 8y + y2 + 25 + 10y = 4y2 + 9 + 12y + y2 + 36 – 12y
⇒ 2y + 29 = 45 ⇒y=8
Hence, coordinates of point P are (16, 8).
13. If two adjacent vertices of a parallelogram are (3, 2) and (– 1, 0) and the diagonals intersect at
(2, – 5), then find the coordinates of the other two vertices. [CBSE (F) 2017]
Sol. Let other two coordinates are (x, y) and (x′, y′).
... O is mid point of AC and BD.
∴ 2 = x+3 D (x', y' ) C (x, y)
2
⇒ x = 1 (2, –5)
O
and, –5 = 2+y
2
⇒ y = –12
Again, –1 + xl = 2 A B
2 (3, 2) (–1, 0)
⇒ x′ = 5 Fig. 6.16
and 0 + yl = –5
2
⇒ y′ = –10
Hence, co-ordinates are (1, –12) and (5, –10).
14. Determine, if the points (1, 5), (2, 3) and (– 2, – 11) are collinear. [NCERT]
Sol. Let A (1, 5), B (2, 3) and C (– 2, – 11) be the given points. Then we have
AB = (2 − 1)2 + (3 − 5)2 = 1 + 4 = 5
BC = (−2 − 2)2 + (−11 − 3)2 = 16 + 196 = 4 × 53 = 2 53
AC = (–2 – 1)2 +(–11 – 5)2 = 9 + 256 = 265
Coordinate Geometry 141
Clearly, AB + BC ≠ AC
\ A, B, C are not collinear.
15. Find the distance between the following pair of points: (a, b), (–a, –b) [NCERT]
Sol. Let two given points be A (a, b) and B(–a, –b).
Here, x1 = a and x2 = –a; y1 = b and y2 = –b
\ AB = (x2 − x1)2 + ( y2 − y1)2
= (−a − a)2 + (−b − b)2 = (−2a)2 + (−2b)2
= 4a2 + 4b2 = 2 a2 + b2 units.
16. Find the area of the triangle whose vertices are : (– 5, – 1), (3, – 5), (5, 2). [NCERT]
Sol. Let A(x1, y1) = (–5, –1), B(x2, y2) = (3, – 5), C(x3, y3) = (5, 2)
\ area of ∆ABC = 1 |x1 (y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)|
2
= 1 |–5(–5 – 2) + 3 (2 + 1) + 5(–1 + 5)|
2
= 1 |35 + 9 + 20| = 1 × 64 = 32 sq units.
2 2
17. If the point A (0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also find the
length of AB. [CBSE Delhi 2014]
Sol. Given that A (0, 2) is equidistant from B (3, p) and C (p, 5).
\ AB = AC
AB2 = AC2
⇒
⇒ (3 – 0)2 + (p – 2)2 = (p – 0)2 + (5 – 2)2
⇒ 32 + p2 + 4 – 4p = p2 + 9
⇒ 4 – 4p = 0
⇒ 4p = 4 ⇒ p = 1
Length of AB = (3 − 0)2 + (1 − 2)2
= 32 + (−1)2 = 9 + 1 = 10 units
18. If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that
bx = ay. [CBSE (AI) 2016]
Sol. Given, PA = PB or (PA)2 = (PB)2
(a + b – x)2 + (b – a – y)2 = (a – b – x)2 + (a + b – y)2
⇒ (a + b)2 + x2 – 2ax – 2bx + (b – a)2 + y2 – 2by + 2ay
= (a – b)2 + x2 – 2ax + 2bx + (a + b)2 + y2 – 2ay – 2by
⇒ 4ay = 4bx or bx = ay
Hence proved.
19. If the point C (–1, 2) divides internally the line segment joining the points A(2, 5) and B(x, y)
in the ratio of 3 : 4, find the value of x2 + y2.
[CBSE (F) 2016]
Sol. ... C (–1, 2) divides AB in the ratio 3 : 4.
∴ 3x + 4(2) = –1 ⇒ x = –5
7 y = –2
and 3y + 4(5) = 2 ⇒ Fig. 6.17
7
\ x2 + y2 = (–5)2 + (–2)2 = 29
142 Xam idea Mathematics–X
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
mathamatics xam idea
Discover the best professional documents and content resources in AnyFlip Document Base.
Search
mathamatics xam idea
- 1 - 50
- 51 - 100
- 101 - 150
- 151 - 200
- 201 - 250
- 251 - 300
- 301 - 350
- 351 - 400
- 401 - 450
- 451 - 500
- 501 - 550
- 551 - 580
Pages: