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 ,(x2 + y2 ), (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, (x2 + y2), (x + y)2 ,....., to n terms.
Here, a = (x - y)2,
d = (x2 + y2) - (x - y)2 = x2 + y2 - x2 - y2 + 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
Here, first term (a) = 126
Common difference (d) = 124 - 126 = -2
(i) an = 3+4n, number of terms = 15
a1 = 3 + 4×1 = 3 + 4 = 7 or a = 7
a2 = 3 + 4×2 = 3 + 8 = 11
d = a2 - a1 = 11 - 7 = 4
(ii) bn = 5 + 2n and number of terms = 15
b1 = 5 + 2×1 = 5 + 2 = 7
b2 = 5 + 2×2 = 5 + 4 = 9
First term (a) = 7 and d = 9 - 7 = 2
(iii) xn = 6 - n and number of terms = 15
x1 = 6 - 1 = 5
x2 = 6 - 2 = 4
First term (a) = 5 and d = 4 - 5 = -1
(iv) yn = 9 - 5n and number of terms = 15
y1 = 9 - 5×1 = 9 - 5 = 4
y2 = 9 - 5×2 = 9 - 10 = -1
First (a) = 4 and d = y2 - y1 = -1 - 4 = -5
6. Find the sum of first 20 terms of the sequence whose nth term is an = An + B.
Solution
nth term = an = 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 an = 2 – 3n.
Solution
8. Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 7 – 3n.
Solution
nth term of an A.P. is an = 7 – 3n and number of terms = 25
Solution
A.P. is given 25, 22, 19,....Sum of the A.P. = 116
Here, a =25 and d = 22 - 25 = 3
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