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

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

Solution

(i) a_n = 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) a_n = 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) a_n= (-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) a_n= 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) a_n= 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) a_n = 5n - 4; a₁₂ and a₁₅
(ii) a_n = (3n - 2)/(4n + 5); a₇ and a₈
(iii) a_n = n(n-1)(n-2); a₅ and a₈
(iv) a_n = (n - 1)(2 - n)(3 + n); a₁₁ a₂₁ a₃
(v) a_n = (-1)ⁿ n; a₃ , a₅ , a₈

Solution

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

(iii) a_n = 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) a_n = (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) a_n = (-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, a_n = a_{n - 1} + 2, n ≥ 2 
(ii) a₁ = a₂ = 2 , a_n = a_{n - 1} - 3, n > 2 
(iii) a₁ = -1, a_n = (a_{n - 1} )/n , n ≥ 2 
(iv) a₁ = 4, a_n = 4 a_{n - 1} + 3, n > 1

Solution

(i) a₁ = 1, a_n = a_n-1 + 2
Let n = 2, 3, 4, 5, 6
a₂ = a_2-1 + 2 = a₁ + 2 = 1+2 = 3 (∵a₁ = 1)
a₃ = a_3-1 + 2 = a₂ + 2 = 3+2 = 5
a₄ = a_4-1 + 2 = a₃ + 2 = 5+2 = 7
a₅ = a_5-1 + 2 = a₄ + 2 = 7+2 = 9
a₆ = a_6-1 + 2 = a₅ + 2 = 9+2 = 11

(ii) a₁ = a2 = 2
an - an-1 - 3
Let n = 3, 4, 5, 6, 7
a₃ = a_3-1 - 3 = a₂ - 3 = 2-3 = -1
a₄ = a_4-1 - 3 = a₃ - 3 = -1-3 = -4
a₅ = a_5-1 - 3 = a₄ - 3 = -4-3 = -7
a₆ = a_6-1 - 3 = a₅ - 3 = -7-3 = -10
a₇ = a_7-1 - 3 = a₆ - 3 = -10-3 = -13

(iv) a₁ = 4, a_n = 4a_n-1 + 3
Let n = 2, 3, 4, 5, 6, then
a₂ = 4a_2-1 + 3 = 4a₁ + 3 = 4×4 + 3 = 16+3 = 19
a₃ = 4a_3-1 + 3 = 4a₂ + 3 = 4×19 + 3 = 76+3 = 79
a₄ = 4a_4-1 + 3 = 4a₃ + 3 = 4×79 + 3 = 316+3 = 319
a₅ = 4a_5-1 + 3 = 4a₄ + 3 = 4×319 + 3 = 1276+3 = 1279
a₆ = 4a_6-1 + 3 = 4a₅ + 3 = 4×1279 + 3 = 5116+3 = 5119