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 nth term are:

(i) an = 3n + 2

(ii) an = (n – 2)/3

(iii) an = 3n

(iv) an = (3n – 2)/ 5

(v) an = (-1)n . 2n

(vi) an = n(n – 2)/2

(vii) an = n2 – n + 1

(viii) an = n2 – n + 1

(ix) an = (2n – 3)/ 6

Solutions:

(i) an = 3n + 2

Given sequence whose an = 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 nth term, an = 3n + 2 are 5, 8, 11, 14, 17.

(ii) an = (n – 2)/3

Given sequence whose R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 1

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

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 3
R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 2

∴ the required first five terms of the sequence whose nth term,

(iii) an = 3n

Given a sequence whose an = 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 nth term, an = 3n are 3, 9, 27, 81, 243.

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 4

(iv) an = (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

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 5

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

(v) an = (-1)n2n

Given sequence whose an = (-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) an = n(n – 2)/2

The given sequence is,
R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 6

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

And, we get

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 7

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

(vii) an = n2 – n + 1

The given sequence whose, an = 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) an = 2n2 – 3n + 1

The given sequence whose an = 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) an = (2n – 3)/ 6

Given the sequence whose,
R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 8

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

And we get

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.1 - 9

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


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