Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 2

Question 14. The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.

Solution: 

Let’s assume that the first term and the common difference of the A.P to be a and d respectively.

Given that,

4th term of an A.P. is three times the first

a₄ = 3(a)

As we know that, to find nth term in an A.P = a + (n – 1)d

a + (4 – 1)d = 3a

3d = 2a ⇒ a = 3d/2 ---------(i)

and,

7th term exceeds twice the third term by 1

a₇ = 2(a3) + 1

a + (7 – 1)d = 2(a + (3–1)d) + 1

a + 6d = 2a + 4d + 1

a – 2d +1 = 0 --------(ii)

Using (i) in (ii), we get

3d/2 – 2d + 1 = 0

3d – 4d + 2 = 0

d = 2

therefore, putting d = 2 in (i), we get a

a = 3

Hence, the first term is 3 and the common difference is 2.

Question 15. Find the second term and the nth term of an A.P. whose 6th term is 12 and the 8th term is 22.

Solution: 

Given that,

a₆ = 12 and a₈ = 22

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore,

a₆ = a + (6-1)d = a + 5d = 12 --------(i)

and,

a₈ = a + (8-1)d = a + 7d = 22 ---------(ii)

Solving (i) and (ii), we get

(ii) – (i)

a + 7d – (a + 5d) = 22 – 12

2d = 10

d = 5

now put d in (i) we get,

a + 5(5) = 12

a = 12 – 25

a = -13

Hence for the A.P: a = -13 and d = 5

so, the nth term is given by an = a + (n-1)d

an = -13 + (n-1)5 = -13 + 5n – 5

an = 5n – 18

Hence, the second term is given by a₂ = 5(2) – 18 = 10 – 18 = -8

Question 16. How many numbers of two digit are divisible by 3?

Solution: 

The first 2 digit number divisible by 3 is 12 and the last 2 digit number divisible by 3 is 99.

this forms an A.P,

12, 15, 18, 21, …., 99

Where, a = 12 and d = 3

now find the number of terms in this A.P

99 = 12 + (n-1)3

99 = 12 + 3n – 3

90 = 3n

n = 90/3 = 30

Hence, there are 30 two digit numbers divisible by 3.

Question 17. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term.

Solution: 

Given that,

An A.P of 60 terms

and a = 7 and a60 = 125

As we know that, to find nth term in an A.P = a + (n – 1)d

a₆₀ = 7 + (60 – 1)d = 125

7 + 59d = 125

59d = 118

d = 2

Therefore, the 32nd term is given by

a₃₂ = 7 + (32 -1)2 = 7 + 62 = 69

Hence a₃₂ = 69

Question 18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.

Solution: 

Given that,

The sum of 4th and 8th terms of an A.P. is 24 and

the sum of the 6th and 10th terms is 34

a₄ + a₈ = 24

and, As we know that, to find nth term in an A.P = a + (n – 1)d

[a + (4-1)d] + [a + (8-1)d] = 24

2a + 10d = 24

a + 5d = 12 ---------(i)

a6 + a10 = 34

[a + 5d] + [a + 9d] = 34

2a + 14d = 34

a + 7d = 17 ----------(ii)

Subtracting (i) form (ii), we get

a + 7d – (a + 5d) = 17 – 12

2d = 5

d = 5/2

Using d in (i) we get,

a + 5(5/2) = 12

a = 12 – 25/2

a = -1/2

Hence, the first term is -1/2 and the common difference is 5/2.

Question 19. The first term of an A.P. is 5 and its 100th term is -292. Find the 50th term of this A.P.

Solution: 

Given that,

a = 5 and a100 = -292

As we know that, to find nth term in an A.P = a + (n – 1)d

a100 = 5 + 99d = -292

99d = -297

d = -3

Hence, the 50th term is

a₅₀ = a + 49d = 5 + 49(-3) = 5 – 147 = -142

Question 20. Find a30 – a20 for the A.P.

(i) -9, -14, -19, -24 (ii) a, a+d, a+2d, a+3d, ……

Solution: 

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore, a₃₀ – a₂₀ = (a + 29d) – (a + 19) =10d

(i) Given A.P. -9, -14, -19, -24

Here, a = -9 and d = -14 – (-9) = = -14 + 9 = -5

therefore, a30 – a20 = 10d

= 10(-5) = -50

(ii) Given A.P. a, a+d, a+2d, a+3d, ……

therefore, a₃₀ – a₂₀ = (a + 29d) – (a + 19d) =10d

Question 21. Write the expression an – ak for the A.P. a, a+d, a+2d, …... Hence, find the common difference of the A.P. for which

(i) 11th term is 5 and 13th term is 79.

(ii) a₁₀ – a₅ = 200

(iii) 20th term is 10 more than the 18th term.

Solution: 

Given A.P. a, a+d, a+2d, …..

therefore, an = a + (n-1)d = a + nd –d

And, ak = a + (k-1)d = a + kd – d

an – ak = (a + nd – d) – (a + kd – d)

= (n – k)d

(i) Given that 11th term is 5 and 13th term is 79,

where n = 13 and k = 11,

a₁₃ – a₁₁ = (13 – 11)d = 2d

79 – 5 = 2d

d = 74/2 = 37

(ii) Given that, a₁₀ – a₅ = 200

(10 – 5)d = 200

5d = 200

d = 40

(iii) Given that, 20th term is 10 more than the 18th term.

a₂₀ – a₁₈ = 10

(20 – 18)d = 10

2d = 10

d = 5

Question 22. Find n if the given value of x is the nth term of the given A.P.

(i) 25, 50, 75, 100, ; x = 1000 (ii) -1, -3, -5, -7, …; x = -151

(iii) 5½, 11, 16½, 22, ….; x = 550 (iv) 1, 21/11, 31/11, 41/11, …; x = 171/11

Solution: 

(i) Given that, A.P. 25, 50, 75, 100, ……,1000

where, a = 25 d = 50 – 25 = 25

nth term= 1000

As we know that, to find nth term in an A.P = a + (n – 1)d

1000 = 25 + (n-1)25

1000 = 25 + 25n – 25

n = 1000/25

n = 40

(ii) Given that, A.P. -1, -3, -5, -7, …., -151

where, a = -1 d = -3 – (-1) = -2

nth term = -151

As we know that, to find nth term in an A.P = a + (n – 1)d

-151 = -1 + (n-1)(-2)

-151 = -1 – 2n + 2

n = 152/2

n = 76

(iii) Given that, A.P. 5½, 11, 16½, 22, …, 550

where, a = 5½ d = 11 – (5½) = 5½ = 11/2

nth term = 550

As we know that, to find nth term in an A.P = a + (n – 1)d

550 = 5½ + (n-1)(11/2)

550 x 2 = 11+ 11n – 11

1100 = 11n

n = 100

(iv) Given that, A.P. 1, 21/11, 31/11, 41/11, 171/11

where, a = 1 d = 21/11 – 1 = 10/11

nth term = 171/11

As we know that, to find nth term in an A.P = a + (n – 1)d

171/11 = 1 + (n-1)10/11

171 = 11 + 10n – 10

n = 170/10

n = 17

Question 23. The eighth term of an A.P is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.

Solution: 

Given that,

a₈ = 1/2(a₂)

a₁₁ = 1/3(a₄) + 1

As we know that, to find nth term in an A.P = a + (n – 1)d

a₈ = 1/2(a₂)

a + 7d = 1/2(a + d)

2a + 14d = a + d

a + 13d = 0 -------(i)

And, a₁₁ = 1/3(a4) + 1

a + 10d = 1/3(a + 3d) + 1

3a + 30d = a + 3d + 3

2a + 27d = 3 --------(ii)

By solving (i) and (ii), by (ii) – 2x(i)

2a + 27d – 2(a + 13d) = 3 – 0

d = 3

Putting d in (i) we get,

a + 13(3) = 0

a = -39

Hence, the 15th term a₁₅ = -39 + 14(3) = -39 + 42 = 3

Question 24. Find the arithmetic progression whose third term is 16 and the seventh term exceeds its fifth term by 12.

Solution:

Given that,

a₃ = 16 and a₇ = a₅ + 12

As we know that, to find nth term in an A.P = a + (n – 1)d

a + 2d = 16…… (i)

and,

a + 6d = a + 4d + 12

2d = 12

d = 6

Using d in (i), we get

a + 2(6) = 16

a = 16 – 12 = 4

Hence, the A.P is 4, 10, 16, 22, …….

Question 25. The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P.

Solution: 

Given that,

a₇ = 32 and a₁₃ = 62

From a_n – a_k = (a + nd – d) – (a + kd – d) = (n – k)d

a₁₃ – a₇ = (13 – 7)d = 62 – 32 = 30

6d = 30

d = 5

Now,

a₇ = a + (7 – 1)5 = 32

a + 30 = 32

a = 2

Hence, the A.P is 2, 7, 12, 17, ……

Question 26. Which term of the A.P. 3, 10, 17, …. will be 84 more than its 13th term ?

Solution: 

Given that,

A.P. 3, 10, 17, ….

where, a = 3 and d = 10 – 3 = 7

a_n = a₁₃ + 84 (Given)

As we know that, to find nth term in an A.P = a + (n – 1)d

3 + (n – 1)7 = 3 + (13 – 1)7 + 84

3 + 7n – 7 = 3 + 84 + 84

7n = 168 + 7

n = 175/7

n = 25

Hence, the 25th term which is 84 more than its 13th term.

Question 27. Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution: 

Let's assume that the two A.Ps be A.P1 and A.P2

For A.P1 the first term is a and the common difference is d

and for A.P2 the first term is b and the common difference is d

Given that,

a100 – b100 = 100

(a + 99d) – (b + 99d) = 100

a – b = 100

Now, the difference between their 1000th terms is,

(a + 999d) – (b + 999d) = a – b = 100

Hence, the difference between their 1000th terms is also 100.