# ICSE Solutions for Chapter 10 Arithmetic Progression Class 10 Mathematics

**Question 1: Find the sum of the first 22 terms of the AP: 8,3.-2**

**Solution 1:**The given A.P. is 8, 3, -2......

Here, a = 8, d = 3 – 8 = - 5 and n = 22

∴ S = n/2[2a + (n - 1)d]

= 22/2 [2×8+ (22 - 1) × (-5)]

= 11[16 + 21×(-5)]

= 11[16 – 105]

= 11 × (-89)

= -979

**Question 2: How many two-digit numbers are divisible by 3?**

**Solution 2:**The two-digit numbers divisible by 3 are as follows:

12, 15, 18, 21, ...., 99

Clearly, this forms an A.P. with first term, a = 12

and common difference, d = 3

Last term = n

^{th}term = 99

The general term of an A.P. is given by

t

_{n}= a + (n - 1)d

⇒ 99 = 12 + (n - 1)(3)

⇒ 99 = 12 + 3n - 3

⇒ 90 = 3n

⇒ n = 30

Thus, 30 two-digit numbers are divisible by 3.

**Question 3: If the p**

^{th}term of an A.P.is (2p+ 3), find the A.P.

**Solution 3:**p

^{th}term of an A.P. = 2p + 3

⇒ t

_{p}= 2p + 3

Putting t = 1, 2, 3, .... we get

t

_{1}= 2×1 + 3 = 2 + 3 = 5

t

_{2}= 2×2 + 3 = 4 + 3 = 7

t

_{3}= 2×3 + 3 = 6 + 3 = 9 and so on.

Thus, the A.P. is 5, 7, 9, …

**Question 4: Find the 24th term of the sequence: 12, 10.8,6.....**

**Solution 4:**

The given sequence is 12, 10, 8, 6,....

Now,

10 - 12 = -2

8 – 10 = -2

6 – 8 = -2, etc.

Hence, the given sequence is an A.P. with first term a = 12

and common difference d = -2.

The general term of an A.P. is given by

t

_{n }= a + (n - 1)d

⇒ t

_{24}= 12 + (24 – 1)(-2) = 12 + 23×(-2) = 12 - 46 = -34

So, the 24

^{th}term is - 34.

**Question 5: If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, Which term of it is zero?**

**Solution 5:**For an A.P.,

t

_{3}= 4

⇒ a + 2d = 4

**...(i)**

t

_{9}= -8

⇒ a + 8d = - 8

**...(ii)**

Subtracting (i) from (ii), we get

6d = -12

⇒ d = -2

Substituting d = -2 in (i), we get

a + 2(-2) = 4

⇒ a – 4 = 4

⇒ a = 8

⇒ General term = t

_{n}= 8 + (n - 1)(-2)

Let p

^{th}term of this A.P. be 0.

⇒ 8 + (p - 1) × (-2) = 0

⇒ 8 – 2p + 2 = 0

⇒ 10 - 2p = 0

⇒ 2p = 10

⇒ p = 5

Thus, 5

^{th}term of this A.P. is 0.

**Question 6: The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms**

**Solution 6:**Sum of first 7 terms of an A.P = 49

⇒ 7/2[2a + 6d] = 49

⇒ 7/2 ×2[a + 3d] = 49

⇒ 7[a + 3d] = 49

⇒ a + 3d = 7

**...(i)**

Sum of first 17 terms of A.P. = 289

⇒ 17/2[2a + 16d] = 289

⇒ 17/2 × 2 [a + 8d] = 289

⇒ 17[a + 8d] = 289

⇒ a + 8d = 17

**... (ii)**

Subtracting (i) from (ii), we get

5d = 10 ⇒ d = 2

Substituting d = 2 in (i), we get

a + 3×2 = 7

⇒ a + 6 = 7 ⇒ a = -1

∴ Sum of first n terms = n/2 [2×1 + (n – 1)2]

= n/2[2 + 2n - 2]

= n/2 × 2n

= n

^{2}

**Question 7: For what value of n, the nth term of A.P 63, 65, 67,.....and nth term of A.P. 3, 10, 17,... are equal to each other?**

**Solution 7:**For an A.P. 63, 65, 67, ....., we have a = 63 and d = 65 – 63 = 2

n

^{th}term = t

_{n}= 63 + (n - 1)×2

For an A.P. 3, 10, 17, ......, we have a' = 3 and d' = 10 – 3 = 7

n

^{th}term = t'

_{n}= 3 + (n - 1) x 7

The two A.P.s will have equal nth terms is

t

_{n}= t’

_{n}

⇒ 63 + (n - 1)×2 = 3 + (n - 1)×7

⇒ 63 + 2n – 2 = 3 + 7n = 7

⇒ 61 + 2n = 7n - 4

⇒ 5n = 65

⇒ n = 13

**Question 8: Is 402 a term of the sequence:**

**8, 13, 18, 23 .......?**

**Solution 8:**

The given sequence is 8, 13, 18, 23,...

Now,

13 – 8 = 5

18 – 13 = 5

23 – 18 = 5, etc

Hence, the given sequence is an A.P. with first term a = 8

and common difference d = 5.

The general term of an A.P. is given by

t

_{n}= a + (n – 1)d

⇒ 402 = 8 + (n - 1)(5)

⇒ 394 = 5n - 5

⇒ 399 = 5n

⇒ n = 399/5

The number of terms cannot be a fraction.

So clearly, 402 is not a term of the given sequence.

**Question 9: If numbers n - 2, 4n - 1 and 5n + 2 are in A.P. find the value of n and its next two terms.**

**Solution 9:**Since (n - 2), (4n - 1) and (5n + 2) are in A.P., we have

(4n - 1) - (n - 2) = (5n + 2) - (4 - 1)

⇒ 4n - 1 – n + 2 = 5n + 2 – 4n + 1

⇒ 3n + 1 = n + 3

⇒ 2n = 2

⇒ n = 1

So, the given numbers are = -1, 3, 7

⇒ a = -1 and d = 3 - (-1) = 4

Hence, the next two terms are (7 + 4) and (7 + 2×4)

i.e. 11 and 15.

**Question 10: In an A.P. the first term is 25, nth term is -17 and the sum of n terms is 132. Find n and the common difference.**

**Solution 10:**First term a = 25

n

^{th}term = -17

⇒ Last term I = -17

Sum of n terms = 132

⇒ n/2 [a + l] = 132

⇒ n(25 - 17) = 264

⇒ n×8 = 264

⇒ n = 33

Now, I = -17

⇒ a + (n - 1)d = -17

⇒ 25 + 32d = -17

⇒ 32d = -42

⇒ d = - 42/32

⇒ d = -21/16

Sum of n terms = 132

⇒ n/2 [a + l] = 132

⇒ n(25 - 17) = 264

⇒ n×8 = 264

⇒ n = 33

Now, I = -17

⇒ a + (n - 1)d = -17

⇒ 25 + 32d = -17

⇒ 32d = -42

⇒ d = - 42/32

⇒ d = -21/16