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.
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 9 |
| Chapter Name | Arithmetic Progressions |
| Exercise | 9.3 |
| Category | RD 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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