# RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progression Exercise 9.1

1. Write the first terms of each of the following sequences whose n th term are

(i) an = 3n + 2
(ii) an = (n - 2)/3
(iii) an = 3n
(iv) an = (3n - 2)/5
(v) an = (-1)n2n
(vi) an = n(n - 2)/2
(vii) an = n2 - n + 1
(viii) an = n2 - n + 1
(ix) an = (2n - 3)/6

Solution

(i) an = 3n + 2
Let n = 1, 2, 3, 4, 5, then
First five terms, a1 = 3×1 + 2 = 3+2 = 5
a2 = 3×2 + 2 = 6+2 = 8
a3 = 3×3 + 2 = 9+2 = 11
a4 = 3×4 + 2 = 12+2 = 14
a5 = 3×5 + 2 = 15+2 = 17

(iii) an = 3n
Let n = 1, 2, 3, 4, 5, then
a1 = 31 = 3
a2 = 32 = 3×3 = 9
a3 = 33 = 3×3×3 = 27
a4 = 34 = 3×3×3×3 = 81
a5 = 35 = 3×3×3×3×3 = 243

(v) an= (-1)n.2n
Let n = 1, 2, 3, 4, 5, then
a1 = (-1)2.21 = -1×2 = -2
a2 = (-1)2.22 = 1×4 = 4
a3 = (-1 )3.23 = -1×8 = -8
a4 = (-1)4.24 = 1×16 = 16
a5 = (-1)5.25 = -1×32 = -32

(vii) an= n2 - n + 1
Let n = 1, 2, 3, 4, 5, then
a1 = 12 -1 + 1 = 1 - 1 + 1 = 1
a2 = 22 - 2 + 1 = 4 - 2 + 1 = 3
a3 = 32 - 3 + 1 = 9 - 3 + 1 = 7
a4 = 42 + 1 = 16 - 4 + 1 = 13
a5 = 52 - 5 + 1 = 25 - 5 + 1 =21

(viii) an= 2n2 - 3n + 1
Let n = 1, 2, 3, 4, 5, then
a1 = 2(1)2 - 3×1 + 1 = 2-3+1 = 0
a2 = 2(2)2 - 3×2 + 1 = 8 - 6 + 1 = 3
a3 = 2(3)2 - 3×3 + 1 = 18 - 9 + 1 = 10
a4 = 2(4)2 - 3×4 + 1 = 32 - 12 + 1 = 21
a5 = 2(5)2 - 3×5 + 1 = 50 - 15 + 1 = 36

2. Find the indicated terms in each of the following sequences whose nth terms are:
(i) an = 5n - 4; a12 and a15
(ii) an = (3n - 2)/(4n + 5); a7 and a8
(iii) an = n(n-1)(n-2); a5 and a8
(iv) an = (n - 1)(2 - n)(3 + n); a11 a21 a3
(v) an = (-1)n n; a3 , a5 , a8

Solution

(i) an = 5n - 4
∴ a12 = 5 × 12 - 4 = 60 - 4 = 56
a13 = 5 × 15 - 4 = 75 - 4 = 71

(iii) an = n(n - 1)(n - 2)
∴ a5 = 5(5 - 1)(5 - 2) = 5 × 4 × 3 = 60
a8 = 8 (8 - 1)(8 - 2) = 8 × 7 × 6 = 336

(iv) an = (n - 1)(2 - n) (3 + n)
a1 = (1 - 1) (2 - 1) (3 + 1)
= 0 × 1 × 4 = 0
a2 = (2 - 1)(2 - 2)(3 + 2)
= 1 × 0 × 5 = 0
a3 = (3 - 1) (2 - 3) (3 + 3)
= 2 × -1 × 6 = -12

(v) an = (-1)nn
a3 = (-1)3.3 = -1 × 3 = -3
a5 = (-1)3.5 = -1 × 5 = -5
a8 = (-1)8.8 = 1 × 8 = 8

3. Find the next five terms of each of the following sequences given by:
(i) a1 = 1, an = an - 1 + 2, n ≥ 2
(ii) a1 = a2 = 2 , an = an - 1 - 3, n > 2
(iii) a1 = -1, an = (an - 1 )/n , n ≥ 2
(iv) a1 = 4, an = 4 an - 1 + 3, n > 1

Solution

(i) a1 = 1, an = an-1 + 2
Let n = 2, 3, 4, 5, 6
a2 = a2-1 + 2 = a1 + 2 = 1+2 = 3 (∵a1 = 1)
a3 = a3-1 + 2 = a2 + 2 = 3+2 = 5
a4 = a4-1 + 2 = a3 + 2 = 5+2 = 7
a5 = a5-1 + 2 = a4 + 2 = 7+2 = 9
a6 = a6-1 + 2 = a5 + 2 = 9+2 = 11

(ii) a1 = a2 = 2
an - an-1 - 3
Let n = 3, 4, 5, 6, 7
a3 = a3-1 - 3 = a2 - 3 = 2-3 = -1
a4 = a4-1 - 3 = a3 - 3 = -1-3 = -4
a5 = a5-1 - 3 = a4 - 3 = -4-3 = -7
a6 = a6-1 - 3 = a5 - 3 = -7-3 = -10
a7 = a7-1 - 3 = a6 - 3 = -10-3 = -13

(iv) a1 = 4, an = 4an-1 + 3
Let n = 2, 3, 4, 5, 6, then
a2 = 4a2-1 + 3 = 4a1 + 3 = 4×4 + 3 = 16+3 = 19
a3 = 4a3-1 + 3 = 4a2 + 3 = 4×19 + 3 = 76+3 = 79
a4 = 4a4-1 + 3 = 4a3 + 3 = 4×79 + 3 = 316+3 = 319
a5 = 4a5-1 + 3 = 4a4 + 3 = 4×319 + 3 = 1276+3 = 1279
a6 = 4a6-1 + 3 = 4a5 + 3 = 4×1279 + 3 = 5116+3 = 5119