# RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progression Exercise 9.6

**1. ****Find the sum of the following arithmetic progressions: **

**(i) 50, 46, 42, ...... to 10 terms (ii) 1, 3, 5, 7, ..... to 12 terms (iii) 3, 9/2, 6, 15/2, ..... to 25 terms (iv) 41, 36, 31, ..... to 12 terms (v) (a + b), (a - b), (a - 3b), ..... to 22 terms (vi ) (x - y)**

^{2},(x

^{2}+ y

^{2}), (x + y)

^{2},......., to n terms (vii) (x - y)/(x + y), (3x - 2y)/(x + y), (5x - 3y)/(x + y), .... to n terms (viii) - 26, -24, -22, ..... to 36 terms

**Solution**

(i) A.P. is 50, 46, 42,.... to 10 terms.

Here, a = 50, d = 46 - 50 = -4 and n = 10

(ii) A.P. is 1, 3, 5, 7,.... to 12 terms.

Here, a = 1, d = 3 - 1 = 2 and n = 12

(iii) A.P. is 3, 9/2, 6, 15/2,.... to 25 terms.

Here, a = 3, d = 9/2 - 3 = 3/2 and n = 25

(iv) A.P. is 41, 36, 31,.... to 12 terms.

Here, a = 41, d = 36 - 41 = -5 and n = 12

(v) A.P. is a+b, a-b, a-3b,.... to 22 terms.

Here, a = a+b, d = a-b - a-b = -2b and n = 22

(vi) A.P. is (x - y)^{2}, (x^{2} + y^{2}), (x + y)^{2} ,....., to n terms.

Here, a = (x - y)^{2},

d = (x^{2} + y^{2}) - (x - y)^{2} = x^{2 }+ y^{2} - x^{2} - y^{2} + 2xy

n = n terms

(viii) A.P. is -26, -24, -22,.... to 36 terms.

**2. Find the sum to n term of the A.P. 5, 2, - 1, -4, -7,......**

**Solution**

**3. Find the sum of n terms of an A.P. whose nth terms is given by an = 5 – 6n.**

**Solution**

^{th}term of the given A.P. is

_{n}= 5 - 6n

_{1}= 5 -6×1 = 5 - 6 = -1

_{2}= 5 - 6×2 = 5 - 12 = -7

_{2}- a

_{1}= -7 - (-1) = -7 + 1 = -6

**4. Find the sum of last ten terms of the A.P.: 8, 10, 12, 14,…, 126.**

**Solution**

Here, first term (a) = 126

Common difference (d) = 124 - 126 = -2

**5. Find the sum of the first 15 terms of each of the following sequences having nth term as**

**(i) a**

_{n}= 3 + 4n**(ii) b**

_{n}= 5 + 2n**(iii) x**

_{n}= 6 – n**(iv) y**

_{n}= 9 – 5n**Solution**

(i) a_{n} = 3+4n, number of terms = 15

a_{1} = 3 + 4×1 = 3 + 4 = 7 or a = 7

a_{2} = 3 + 4×2 = 3 + 8 = 11

d = a_{2} - a_{1} = 11 - 7 = 4

(ii) b_{n} = 5 + 2n and number of terms = 15

b_{1} = 5 + 2×1 = 5 + 2 = 7

b_{2} = 5 + 2×2 = 5 + 4 = 9

First term (a) = 7 and d = 9 - 7 = 2

(iii) x_{n} = 6 - n and number of terms = 15

x_{1} = 6 - 1 = 5

x_{2} = 6 - 2 = 4

First term (a) = 5 and d = 4 - 5 = -1

(iv) y_{n} = 9 - 5n and number of terms = 15

y_{1} = 9 - 5×1 = 9 - 5 = 4

y_{2} = 9 - 5×2 = 9 - 10 = -1

First (a) = 4 and d = y_{2} - y_{1} = -1 - 4 = -5

**6. Find the sum of first 20 terms of the sequence whose nth term is a _{n} = An + B.**

**Solution**

nth term = a_{n} = An + B and number of terms = 20

= 10[2A + 2B + 19A]

= 10[21A + 2B

= 210A + 20B

**7. Find the sum of the first 25 terms of an A.P. whose nth term is given by a _{n} = 2 – 3n.**

**Solution**

**8. Find the sum of the first 25 terms of an A.P. whose nth term is given by a _{n} = 7 – 3n.**

**Solution**

_{n}= 7 – 3n and number of terms = 25

**9. If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, …, is 116. Find the last term.**

**Solution**

Sum of the A.P. = 116

Here, a =25 and d = 22 - 25 = 3

**10. (i) How many terms of the sequence 18, 16, 14, … should be taken so that their sum is zero ?**

**(ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?**

**(iii) How many terms of the A.P. 9, 17, 25,… must be taken so that their sum is 636 ?**

**(iv) How many terms of the A.P. 63, 60, 57, ……… must be taken so that their sum is 693 ?**

**(v) How many terms of the A.P. 27, 24, 21, …, should be taken so that their sum is zero?**

**Solution**

(i) In the sequence 18, 16, 14,....

Sum = 0

Here, first term (a) = 18 and common difference (d) = 16 - 18 = -2

Let n be the number of terms

Either x = 0, which is not possible.

or, 38 - 2n = 0, then 2n = 38

⇒ n = 19

Number of terms in the sequence = 19.

(ii) First term of an A.P. = -14

and fifth term =2

Sum of terms = 40

Let n be the number of terms, then

Either n - 10 = 0, then n = 10

or, n + 2 = 0

⇒ n = -2 but it is not possible being negative.

Number of terms = 10

(iii) A.P. is 9, 17, 25,....

Let number of terms of the A.P. = n

First term (a) = 9

Common difference (d) = 17 - 9 = 8 and sum of n terms = 636

(iv) Given A.P. = 63, 60, 57,....

Sum = 693

Let number of terms = n

First term (a) = 63

Common difference (d) = 60 - 63 = -3

Either n - 21 = 0, then n = 21

or, n - 22 = 0, then n = 22

Number of terms = 21 or 22.

(v) A.P. = 27, 24, 21,....

a = 27

d = 24 - 27 = -3

Sum = 0

Let n terms be there in A.P.

**11. Find the sum of the first(i) 11 terms of the A.P. : 2, 6, 10, 14,…(ii) 13 terms of the A.P. : -6, 0, 6, 12,…(iii) 51 terms of the A.P.: whose second term is 2 and fourth term is 8.**

**Solution**

(i) A.P. is 2, 6, 10, 14,.... in which first term(a) = 2 and common difference(d) = 6 - 2 = 4

(ii) A.P. is 2, 6, 10, 14,.... in which first term(a) = -6 and common difference(d) = 0 - (-6) = 6

(iii) Second term of an A.P. = 2 and fourth term = 8

Let a be the first term and de be the common difference.

**12. Find the sum of(i) the first 15 multiples of 8(ii) the first 40 positive integers divisible by(a) 3, (b) 5, (c) 6(iii) all 3-digit natural numbers which are divisible by 13.(iv) all 3-digit natural numbers, which are multiples of 11.(v) all 2-digit natural numbers divisible by 4. **

**Solution**

(i) First 15 multiples of 8 are 8, 16, 24, 32, ..., 120 in which

first term (a) = 8

common difference (d) = 8

and last term = 120

(ii) (a) First 40 positive integers divisible by 3 are 3, 6, 9, 12, 15, ..., 120 in which

a = 3,

d = 3

and l = 120

(b) 40 multiple of 5 are 5, 10, 15, 25, ... 200 in which

first term (a) = 5

common difference (d) = 5

and last term (l) = 200

(c) 40 multiples of 6 are 6, 12, 18, 24, ... 240 in which

first term (a) = 6

common difference (d) = 6

(iii) 3-digit numbers which are divisible by 13 will be 104, 117, 130, 143, ... 988

Here, first term (a) = 104 Common difference (d) = 13 and last term (1) = 988

Let n be the numbers of terms.

(iv) 3-digit numbers which are multiple of 11 are 110, 121, 132, 143, 154, 165, 176, 187, 198, ..., 990

S = 110 + 121 + 132 + 143 +...+ 990

= 11[10+11+12 +13 +...+ 90]

Here, a = 10, d = 1, l = 90

Last term = a + (n - 1) d

90 = 10 + (n - 1)×1

⇒ 90 = 10 + n - 1

⇒ n = 90 - 10 + 1 = 81

⇒ n = 81

(v) The two digits numbers which are divisible by 4 are 12, 16, 20, ..., 96 are in A.P.

Here, a = 12 and d = 4