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

Sequences are observed all around us in everyday life. In this chapter, we are specifically concentrating on one type of sequence known as arithmetic progression. The first exercise of this Chapter 9 is an introductory exercise to writing down terms of a sequence. For a clear understanding of these basics, you can access the RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.1 PDF provided below for detailed solutions to problems.

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

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Access RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions 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)ⁿ . 2ⁿ

(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

Solutions:

(i) a_n = 3n + 2

Given sequence whose a_n = 3n + 2

To get the first five terms of a given sequence, put n = 1, 2, 3, 4, 5, and we get

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

∴ the required first five terms of the sequence whose n^th term, a_n = 3n + 2 are 5, 8, 11, 14, 17.

(ii) a_n = (n – 2)/3

Given sequence whose

On putting n = 1, 2, 3, 4, 5, then can get the first five terms

∴ the required first five terms of the sequence whose n^th term,

(iii) a_n = 3ⁿ

Given a sequence whose a_n = 3ⁿ

To get the first five terms of a given sequence, put n = 1, 2, 3, 4, 5 in the above

a₁ = 3¹ = 3;

a₂ = 3² = 9;

a₃ = 27;

a₄ = 3⁴ = 81;

a₅ = 3⁵ = 243.

∴ the required first five terms of the sequence whose n^th term, a_n = 3ⁿ are 3, 9, 27, 81, 243.

(iv) a_n = (3n – 2)/ 5

Given sequence whose

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above

And, we get

∴ the required first five terms of the sequence are 1/5, 4/5, 7/5, 10/5, 13/5

(v) a_n = (-1)ⁿ2ⁿ

Given sequence whose a_n = (-1)ⁿ2ⁿ

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above.

a₁ = (-1)¹.2¹ = (-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

∴ the first five terms of the sequence are – 2, 4, – 8, 16, – 32.

(vi) a_n = n(n – 2)/2

The given sequence is,

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.

And, we get

∴ the required first five terms are -1/2, 0, 3/2, 4, 15/2

(vii) a_n = n² – n + 1

The given sequence whose, a_n = n² – n + 1

To get the first five terms of a given sequence, put n = 1, 2, 3, 4, 5.

And we get

a₁ = 1² – 1 + 1 = 1

a₂ = 2² – 2 + 1 = 3

a₃ = 3² – 3 + 1 = 7

a₄ = 4² – 4 + 1 = 13

a₅ = 5² – 5 + 1 = 21

∴ the required first five terms of the sequence are 1, 3, 7, 13, 21.

(viii) a_n = 2n² – 3n + 1

The given sequence whose a_n = 2n² – 3n + 1

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.

And we get

a₁ = 2.1² – 3.1 + 1 = 2 – 3 + 1 = 0

a₂ = 2.2² – 3.2 + 1 = 8 – 6 + 1 = 3

a₃ = 2.3² – 3.3 + 1 = 18 – 9 + 1 = 10

a₄ = 2.4² – 3.4 + 1 = 32 – 12 + 1 = 21

a₅ = 2.5² – 3.5 + 1 = 50 – 15 + 1 = 36

∴ the required first five terms of the sequence are 0, 3, 10, 21, 36.

(ix) a_n = (2n – 3)/ 6

Given the sequence whose,

To get the first five terms of the sequence, we put n = 1, 2, 3, 4, 5.

And we get

∴ the required first five terms of the sequence are -1/6, 1/6, 1/2, 5/6 and 7/6