RD Sharma Class 10 Ex 9.2 Solutions Chapter 9 Arithmetic Progressions

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 9
Chapter NameArithmetic Progressions
Exercise9.2
CategoryRD Sharma Solutions

RD Sharma Solutions for Class 10 Chapter 9 Arithmetic Progressions Ex 9.2 Download PDF

Chapter 9: Arithmetic Progressions Exercise 9.2

Question: 1

For the following arithmetic progressions write the first term a and the common difference d:

(i) – 5, -1, 3, 7, ………

(ii) 1/5, 3/5, 5/5, 7/5, ………

(iii) 0.3, 0.55, 0.80, 1.05, …….. 

(iv) -1.1, – 3.1, – 5.1, – 7.1, ………. 

Solution:

We know that if a is the first term and d is the common difference, the arithmetic progression is a, a + d, a + 2d + a + 3d, ………… 

(i) – 5, –1, 3, 7, ………… 

Given arithmetic series is – 5, –1, 3, 7 …………. 

This is in the form of a, a + d, a + 2d + a + 3d, ………… by comparing these two a = – 5, a + d = 1, a + 2d = 3, a + 3d = 7, 

First term (a) = – 5

By subtracting second and first term, we get

(a + d) – (a) = d

-1 – (- 5) = d

4 = d

Common difference (d) = 4.

(ii) 1/5, 3/5, 5/5, 7/5, ………….

Given arithmetic series is,

1/5, 3/5, 5/5, 7/5, ……………

This is in the form of 1/5, 2/5, 5/5, 7/5, ……….. a, a + d, a + 2d, a + 3d,

By comparing this two, we get

a = 1/5, a + d = 3/5, a + 2d = 5/5, a + 3d = 7/5

First term cos = 1/5

By subtracting first term from second term, we get

d = (a + d)-(a)

d = 3/5 – 1/5

d = 2/5

common difference (d) = 2/5

(iii) 0.3, 0.55, 0.80, 1.05, ………… 

Given arithmetic series,

0.3, 0.55, 0.80, 1.05, ………. 

General arithmetic series

a, a + d, a + 2d, a + 3d, 

By comparing,

a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05

First term (a) = 0.3.

By subtracting first term from second term. We get

d = (a + d) – (a)

d = 0.55 – 0.3

d = 0.25

Common difference (d) = 0.25

(iv) –1.1, – 3.1, – 5.1, –7.1, ………….

General series is

a, a + d, a + 2d, a + 3d, …………….

By comparing this two, we get

a = –1.1, a + d = –3.1, a + 2d = –5.1, a + 3d = –71

First term (a) = –1.1

Common difference (d) = (a + d) – (a)

= -3.1 – ( – 1.1)

Common difference (d) = – 2

Question: 2

Write the arithmetic progressions write first term a and common difference d are as follows:

(i) a = 4, d = – 3

(ii) a = –1, d = 1/2

(iii) a = –1.5, d = – 0.5

Solution:

We know that, if first term (a) = a and common difference = d, then the arithmetic series is, a, a + d, a + 2d, a + 3d, 

(i) a = 4, d = -3

Given first term (a) = 4

Common difference (d) = -3

Then arithmetic progression is, a, a + d, a + 2d, a + 3d, …………..

⟹ 4, 4 – 3, a + 2(-3), 4 + 3(-3), ………….. 

⟹ 4, 1, – 2, – 5, – 8 ……………. 

(ii) a = -1, d = 1/2

Given, First term (a) = -1

Common difference (d) = 1/2

Then arithmetic progression is,

⟹ a, a + d, a + 2d, a + 3d, 

⟹ -1, -1 + 1/2, -1, 2½, -1 + 3½, ………..  

⟹ -1, -1/2, 0, 1/2

(iii) a = –1.5, d = – 0.5

Given First term (a) = –1.5

Common difference (d) = – 0 5

Then arithmetic progression is

⟹ a, a + d, a + 2d, a + 3d, …………… 

⟹ -1.5, -1.5, -0.5, –1.5 + 2(– 0.5), –1.5 + 3(– 0.5)

⟹ – 1.5, – 2, – 2.5, – 3, …………….

Then required progression is

-1.5, – 2, – 2.5, – 3, …………… 

Question: 3

In which of the following situations, the sequence of numbers formed will form an A.P.? 
(i) The cost of digging a well for the first metre is Rs 150 and rises by Its 20 for each succeeding metre. 
(ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder. 

Solution:

(i) Given,

Cost of digging a well for the first meter (c1) = Rs.150.

Cost rises by Rs.20 for each succeeding meter

Then,

Cost of digging for the second meter (c2) = Rs.150 + Rs 20 = Rs 170

Cost of digging for the third meter (c3) = Rs.170 + Rs 20 = Rs 210

Thus, costs of digging a well for different lengths are 150, 170, 190, 210, ……….. 

Clearly, this series is in A∙P.

With first term (a) = 150, common difference (d) = 20

(ii) Given

Let the initial volume of air in a cylinder be V liters each time 3th/4 of air in a remaining 1.e

1 -1/4

First time, the air in cylinder is 3/4 V. 

Second time, the air in cylinder is 3/4 V. 

Third time, the air in cylinder is (3/4)2 V.

Therefore, series is V, 3/4 V, (3/4)V,(3/4)V, ……..

Question: 4

Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference.

Solution:

Given sequence is

an = 5n – 7

nth term of given sequence (an) = 5n -7

(n + 1)th term of given sequence (an + 1) – an

= (5n – 2) – (5n – 7) = 5

∴ d = 5

Question: 5

Show that the sequence defined by an = 3n2 – 5 is not an A.P.

Solution:

Given sequence is,

an = 3n2 – 5.

nth term of given sequence (an) = 3n2 – 5.

(n + 1)th term of given sequence (an + 1) = 3(n + 1)2 – 5

= 3(n2 + 12 + 2n.1) – 5

= 3n2 + 6n -2

∴ The common difference (d) = an + 1 – an

d = (3n2 + 6n – 2) – (3n2 – 5)

= 3a2 + 6n – 2 – 3n2 + 5

= 6n + 3

Common difference (d) depends on ‘n’ value

∴ given sequence is not in A.P. 

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