RD Sharma Class 10 Ex 9.3 Solutions Chapter 9 Arithmetic Progressions

In this chapter, we provide RD Sharma Class 10 Ex 9.3 Solutions Chapter 9 Arithmetic Progressions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 9.3 Solutions Chapter 9 Arithmetic Progressions pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 9
Chapter NameArithmetic Progressions
Exercise9.3
CategoryRD Sharma Solutions

RD Sharma Solutions for Class 10 Chapter 9 Arithmetic Progressions Ex 9.3 Download PDF

Chapter 9: Arithmetic Progressions Exercise 9.3

Question: 1

Find:

(i) 10th tent of the AP 1, 4, 7, 10….

(ii) 18th term of the AP √2, 3√2, 5√2, ……. 

(iii) nth term of the AP 13, 8, 3, -2, ………. 

(iv) 10th term of the AP -40, -15, 10, 35, …………. 

(v) 8th term of the AP 11, 104, 91, 78, …………… 

(vi) 11th tenor of the AP 10.0, 10.5, 11.0, 11.2, ………….. 

(vii) 9th term of the AP 3/4, 5/4, 7/4 + 9/4, ………..

Solution:

Given A.P. is 1, 4, 7, 10, ………. 

First term (a) = 1

Common difference (d) = second than first term

= 4 – 1 = 3.

nth term in an A.p = a + (n – 1)d

10th term in an 1 + (10 – 1)3

= 1 + 9.3

= 1 + 27

= 28

(ii) Given A.P. is

√2, 3√2, 5√2, …….

Fiat term (a) = √2 

Common difference = Second term – First term

= 3√2 – √2

d = 2√2 

nth term in an A. P. = a + (n – 1)d

18th term of A. P. = √2 + (18 – 1)2√2

= √2 + 17.2√2

= √2 (1+34)

= 35√2

∴ 18th term of A. P. is 35√2

(iii) Given A. P. is

13, 8, 3, – 2,  ………… 

First term (a) = 13

Common difference (d) = Second term first term

= 8 – 13 = – 5

nth term of an A.P. an = a +(n – 1)d

= 13 + (n – 1) – 5

= 13 – 5n + 5

an = 18 – 5n

(iv) Given A. P. is

– 40, -15, 10, 35, ………. 

First term (a) = -40

Common difference (d) = Second term – fast term

= -15 – (- 40)

= 40 – 15

= 25

nth term of an A.P. an = a + (n – 1)d

10th term of A. P. a10 = -40 + (10 – 1)25

= – 40 + 9.25

= – 40 + 225

= 185

(v) Given sequence is 117, 104, 91, 78, …………. 

First learn can = 117

Common difference (d) = Second term – first term

= 104 – 117

= – 13

nth term = a + (n – 1)d

8th term = a + (8 – 1)d

= 117 + 7(-13)

= 117 – 91

= 26

(vi) Given A. P is

10.0, 10.5, 11.0, 11.5, 

First term (a) = 10.0

Common difference (d) = Second term – first term

= 10.5 – 10.0 = 0.5

nth term; an = a + (n – 1)d

11th term a11 = 10.0 + (11 – 1)0.5

= 10.0 + 10 x 0.5

= 10.0 + 5

=15.0

(vii) Given A. P is

3/4,  5/4, 7/4, 9/4, …………

First term (a) = 3/4

Common difference (d) = Second term – first term

= 5/4 – 3/4

= 2/4

nth term an = a + (n – 1)d

9th term a9 = a + (9 – 1)d

Question: 2

(i) Which term of the AP 3, 8, 13, …. is 248?

(ii) Which term of the AP 84, 80, 76, … is 0?

(iii) Which term of the AP 4. 9, 14, …. is 254?

(iv) Which term of the AP 21. 42, 63, 84, … is 420?

(v) Which term of the AP 121, 117. 113, … is its first negative term?

Solution:

(i) Given A.P. is 3, 8, 13, ………..

First term (a) = 3

Common difference (d) = Second term – first term

= 8 – 3

= 5

nth term (an) = a + (n – 1)d

Given nth term an = 248

248 = 3+(n – 1).5

248 = -2 + 5n

5n = 250

n =250/5 = 50

50th term is 248.

(ii) Given A. P is 84, 80, 76, ………… 

First term (a) = 84

Common difference (d) = a2 – a

= 80 – 84

= – 4

nth term (an) = a +(n – 1)d

Given nth term is 0

0 = 84 + (n – 1) – 4

84 = +4(n –  1)

n – 1 = 84/4 = 21

n = 21 + 1 = 22

22nd term is 0.

(iii) Given A. P 4, 9, 14, ………… 

Fiat term (a) = 4

Common difference (d) = a2 – a

= 9 – 4

= 5

nth term (an) = a + (n – 1)d

Given nth term is 254

4 + (n – 1)5 = 254

(n – 1)∙5 = 250

n – 1 = 250/5 = 50 

n = 51

∴  51th term is 254. 

(iv) Given A. P

21, 42, 63, 84, ……… 

a = 21, d = a2 – a

= 42 – 21

= 21

nth term (an) = a +(n – 1)d

Given nth term = 420

21 + (n – 1)21 = 420

(n – 1)21 = 399

n – 1 = 399/21 = 19

n = 20

∴ 20th term is 420.

(v) Given A.P is 121, 117, 113, ……….. 

Fiat term (a) = 121

Common difference (d) = 117 – 121

= – 4

nth term (a) = a + (n – 1)d

Given nth term is negative i.e., an < 0

121 + (n – 1) – 4 < 0

121 + 4 – 4n < 0

125 – 4n < 0

4n > 125

n > 125/4

n > 31.25

The integer which comes after 31.25 is 32.

∴ 32nd term is first negative term

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