**RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progression** Ex 5.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.2.

## Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.2

Ex 5.2 Class 10 RBSE Question 1.

Find

(i) 10^{th} term of A.P. 2, 7, 12, ….

(ii) 18^{th} term of A.P. √2, 3√2, 5√2, …..

(iii) 24^{th} term of A.P. 9, 13, 17, 21, ….

Solution :

(i) Given A.P. 2, 7, 12, …..

First term a = 2

Common (RBSESolutions.com) difference d = 7 – 2 = 5

n^{th} term a_{n} = a + (n – 1)d

∴ a_{10} = 2 + (10 – 1) × 5

= 2 + 9 × 5

= 2 + 45

= 47

Hence, 10^{th} term of given series 47.

(ii) Given A.P. √2, 3√2, 5√2, …..

First term a = √2

Common term d = 3√2 – √2

= √2(3 – 1) = 2√2

n^{th} term a_{n} = a + (n – 1)d

∴ a_{18} = √2 + (18 – 1)2√2

= √2 + 17 × 2√2

= √2 + 34√2

= 35√2

Hence a_{18} = 35√2

(iii) Given A.P. 9, 13, 17, 21, …..

First term a = 9

common (RBSESolutions.com) difference

d = 13 – 9 = 4

n^{th} term a_{n} = a + (n – 1)d

∴ a_{24} = 9 + (24 – 1) × 4

= 9 + 23 × 4

= 9 + 92

= 101

Hence a_{24} = 101

RBSE Solutions For Class 10 Maths Chapter 5.2 Question 2.

Solve

(i) Which term of A.P. 21, 18, 15, …… is – 81?

(ii) Which term of AP. 84, 80, 76, …. is zero ?

(iii) Is 301 any term of series 5, 11, 17, 23, …… ?

(iv) Is -150 is any term of A.P. 11, 8, 5, 2, …….

Solution :

(i) Given A.P. 21, 18, 15, ….

First term (a) = 21

Common (RBSESolutions.com) difference

(d) = 18 – 21 = -3

∵ a_{n} = a + (n – 1)d

According to question

-81 = 21 + (n – 1)(-3)

⇒ -81-21 = (n – 1) × -3

⇒ -102 = (n – 1) × -3

⇒ (n – 1) = \(\frac { -102 }{ -3 }\)

⇒ n – 1 = 34

⇒ n = 34 + 1 = 35

Hence, 35^{th} term of given series is -81.

(ii) Given A.P. 84, 80, 76…..

First term (a) = 84

Common (RBSESolutions.com) difference (d) = 80 – 84 = -4

∵ a_{n} = a + (n – 1)d

According to question,

0 = 84 + (n – 1)(-4)

⇒ -84 = (n – 1) × -4

⇒ (n – 1) = \(\frac { -84 }{ -4 }\)

⇒ n – 1 = 21

⇒ n = 21 + 1 = 22

Hence. 22^{nd} term of given A.P. is zero.

(iii) Given A.P. 5, 11, 17, 23 …..

First term (a) = 5

Common difference

(d) = 11 – 5 = 6

∵ a_{n} = a + (n – 1)d

According to (RBSESolutions.com) question.,

⇒ 301 = 5 + (n – 1)(6)

⇒ 301 – 5 = 6(n—1)

⇒ 6(n – 1) = 296

⇒ (n – 1) = \(\frac { 296 }{ 6 }\)

⇒ n – 1 = 49.33

⇒ n = 49.33 + 1 = 50.33

∴ Value of n cannot be fraction, it means n is not a whole number. Thus no term can be 301 in given series A.P.

(iv) Given A.P. : 11, 8, 5, 2 …

First term (a) = 11 and common difference (d) = 8 – 11 = – 3

Let n^{th} term, a_{n} = -150

⇒ a + (n – 1)d = -150

⇒ 11 + (n – 1) × (-3) = -150

⇒ -3(n – 1) = -150 – 11 = -161

⇒ (n – 1) = \(\frac { -161 }{ -3 }\)

= 53.6 (approx)

∴ n = 53.6 + 1 = 54.6

⇒ n is not a whole no.

Hence, -150 is no (RBSESolutions.com) term is given A.P.

Exercise 5.2 Class 10 RBSE Question 3.

If 6^{th} term and 17th turn of A.P. are 19 and 41 respectively, then find 40^{th} term.

Solution :

Given

6^{th} term a_{6} = 19

and 17^{th} term a_{17} = 41

40^{th} term a_{40} = ?

n^{th} term, a = a + (n – 1)d

a_{6} = a + (6 – 1)d

⇒ 19 = a + 5d …..(i)

and a_{17} = a + (17 – 1)d

⇒ 41 = a + 16 d …(ii)

On subtracting (i) from (ii)

d = 2

Putting d = 2 in (RBSESolutions.com) equation (i)

a + 5 × 2 = 19

a = 19 – 10

a = 9

Thus, a_{40} = a + (40 – 1)d

= 9 + 39 × 2

= 9 + 78 = 87

Hence, 40^{th} term of A.P. is 87.

RBSE Solutions For Class 10 Maths Chapter 5 Question 4.

Third and ninth term of an A.P. (RBSESolutions.com) are 4 and -8 respectively, then its which term will be zero?

Solution :

Let a is first term of A.P. and d is common difference.

Given, third term a_{3} = 4

a + (3 – 1)d = 4, [a_{n} = a + (n – 1)d]

⇒ a + 2d = 4

and ninth term a_{9} = -8

a + (9 – 1)d = -8

⇒ a + 8d = -8 …(ii)

Subtracting equation (i) from (ii)

d = \(\frac { -22 }{ -11 }\)

∴ d = \(\frac { -12 }{ 6 }\) = -2

Putting this value (RBSESolutions.com) of d in equation (i)

a + 2(-2) = 4

or a – 4 = 4

∴ a = 4 + 4 = 8

Let n^{th} term of series will be zero, than n^{th} term

n^{th} term a_{n} = 0

∴ a + (n – 1)d = 0

⇒ 8 + (n – 1) × (-2) = 0

⇒ -2(n – 1) = -8

⇒ (n – 1) = 4

∴ n = 5

Hence, 5^{th} term of AP. will be zero

RBSE Class 10 Maths Chapter 5 Question 5.

Third term of an A.P. Is 16 and 7^{th} term is 12 more than (RBSESolutions.com) 5^{th} term, then find AP.

Solution :

Let first term of A.P. is a and common difference is d.

Given a_{3} = 16

a + (3 – 1)d = 16

⇒ a + 2d = 16 …(i)

According to question, a_{7} = 12 – a_{5}

⇒ a_{7} – a_{5} = 12

[a + (7 – 1)d] – [a+(5 – 1)d] = 12

⇒ a_{7} + 6d – a – 4d = 12

⇒ a + 6d – a – 4d = 12

⇒ 2d = 12

d = \(\frac { 12 }{ 2 }\) = 6

Substituting value of d in equation (i)

a + 2(6) = 16

∴ a = 16 – 12 = 4

AP. a, a + d, a + 2d,…

= 4, 4 + 6, 4 + 2 × 6,…

= 4, 10, 16,…

Hence, required A.P. is 4, 10, 16, 22,…

RBSE Class 10 Maths Chapter 5.2 Question 6.

How many three digit (RBSESolutions.com) numbers are divisible by 7 ?

Solution :

Series of 3 digit numbers 100, 101, 102, …….. 999,

First three digit number divisible by 7 = 105 and last number = 994

Then series of three digit numbers divisible by 7

105, (105 + 7), (105 + 7+ 7), …. 994 = 105, 112, 119, …,994

Let total number of terms is n.

First term a = 105, common difference d = 7,

∴ n^{th} term a_{n} = 994

⇒ a + (n – 1)d = 994

⇒ 105 + (n – 1) × 7 = 994

⇒ (n – 1) × 7=994 – 105 = 889

⇒ (n – 1) = \(\frac { 889 }{ 7 }\) = 127

∴ n = 127 +1 = 128

Hence, three digit number divisible by 7 is 128.

Class 10 Maths RBSE Solution Chapter 5 Question 7.

Find 11^{th} term from (RBSESolutions.com) last of A.P. 10, 7, 4, …., -62

Solution :

Given A.P. 10, 7, 4….., -62

First term (a) = 10

Common difference (d) = 7 – 10 = – 3

Last term a_{n} = -62

Formula, r^{th} term from last

= a_{n} – (r – 1)d

11^{th} term from last = -62 – (11 – 1) × (-3)

= -62 – 10(-3)

= -62 + 30

= -32

Hence, 11^{th} term from last of A.P. = 32.

RBSE Class 10 Maths Chapter 5 Exercise 5.2 Question 8.

Find the 12^{th} term from (RBSESolutions.com) last of A.P. 1, 4, 7, 10, …., 88

Solution :

Given A.P. 1, 4, 7, 10……, 88

First term (a) = 1

Common difference (d) = 4 – 1 = 3

Last term a_{n} = 88

Formula, r^{th} term from last

= a_{n} – (r – 1)d

12^{th} term from last = 88 – (12 – 1) × 3

= 88 – 11 × 3

= 88 – 33 = 55

Hence, 12^{th} term from last term of A.P will be 55.

Chapter 5 Class 10 Maths RBSE Question 9.

There are 60 terms in an AP. If its first and (RBSESolutions.com) last term are 7 and 125 respectively, then find its 32^{nd} term.

Solution :

Number of terms (n) = 60

First term (a) = 7

Last term (a_{n}) = 125

Formula, a_{n} = a + (n – 1)d

⇒ 125 = 7 + (60 – 1)d

⇒ 125 – 7 = 59 d

⇒ 118 = 59 d

⇒ d = \(\frac { 118 }{ 59 }\)

⇒ d = 2

Thus, 32^{nd} term

a_{32} = a + (32 – 1)d

= 7 + 31 × 2

= 7 + 62 = 69

Hence, 32^{nd} term of A.P. is 69.

RBSE Class 10 Chapter 5 Question 10.

Four numbers are ¡n A.P. If sum of (RBSESolutions.com) numbers is 50 and larger number is 4 times the smaller number, then find the number.

Solution :

Let four numbers in A.P. are

a, a + d, a + 2d, a + 3d

According to question

a + (a + d) + (a + 2d) + (a + 3d) = 50

⇒ 4a + 6d = 50

⇒ 2(2a + 3d) = 50

⇒ 2a + 3d = 25 …..(i)

If larger number is 4 times the smaller (RBSESolutions.com) number, then equation will be as follow,

a + 3d = 4 x a

⇒ a + 3d = 4a

⇒ 3d = 3a

⇒ d = a …(ii)

from equation (i) and (ii)

2a + 3a = 25 [∴ d = a]

a = 25

a = 5

∴ d = 5 [∵ eq^{n}(ii)]

∴ Numbers a = 5

a + d = 5 + 5 = 10

a + 2d = 5 + 2 × 5 = 15

a + 3d = 5 + 3 × 5 = 20

Hence, four numbers are 5, 10, 15, and 20

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