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

## Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Miscellaneous Exercise

Question 1.

The common difference of two A.P.’s are same. (RBSESolutions.com) First term of one such in 8 and of other is 3. Then difference between their 30^{th} terms :

(A) 11

(B) 3

(C) 8

(D) 8

Solution :

30^{th} term of first A.P.

a_{30} = 8 + (30 – 1)d

= 8 + 29d

30^{th} term of second A.P.

a’_{30} = 3 + (30 – 1)d

= 3 + 29d

a_{30} – a’_{30} = 5

Hence, option (D) is correct.

Question 2.

If 18, a, b, -3 are in A.P. (RBSESolutions.com) then a + b =

(A) 19

(B) 7

(C) 11

(D) 15

Solution :

First term = 18

Second term 18 + d = a

Third term 18 + 2d = b

Fourth term 18 + 3d = -3

⇒ 3d = -3 – 18

⇒ 3d = -21

⇒ d = \(\frac { -21 }{ 3 }\) = -7

∴ Second term a = 18 + (-7) = 11

Third term b = 18 + 2 × (-7)

= 18 – 14 = 4

Thus, a + b = 11 + 4 = 15

Hence option (D) is correct

Question 3.

If 7^{th} and 13^{th} term of A.P. are 34 and 64 (RBSESolutions.com) respectively. Then its 18^{th} term is :

(A) 89

(B) 88

(C) 87

(D) 90

Solution :

a_{7} = 34 (given)

a_{13} = 64 (given)

– 6d = -30

d = 5

put the value of d in a + 6d = 34

a + 6 × 5 = 34

a = 34 – 30

a = 4

∴ a_{18} = a + 17d

= 4 + 17 × 5 = 4 + 85 = 89

Thus, option (A) is correct.

Question 4.

First and last term of A.P. are 2 and 34 respectively. (RBSESolutions.com) Sum of its terms is 90 then value of n will be :

(A) 3

(B) 4

(C) 5

(D) 6

Solution :

We have a = 2, l = 34, d = 8, S_{n} = 90

⇒ S_{n} = \(\frac { n }{ 2 }\) (a + l)

⇒ 90 = \(\frac { n }{ 2 }\) (a + l)

⇒ \(\frac { n }{ 2 }\) × 36 = 90

⇒ n × 18 = 90

⇒ n = \(\frac { 90 }{ 18 }\) = 5

⇒ n = 5

Hence, option (C) is correct.

Question 5.

If sum of n terms of A.P. is 3n^{2} + 5n, then its (RBSESolutions.com) which term is 164 :

(A) 12^{th}

(B) 15^{th}

(C) 27^{th}

(D) 20^{th}

Solution :

Given : S_{n} = 3n^{2} + 5n

S_{1} = 3(1)^{2} + 5(1) = 8

S_{2} = 3(2)^{2} + 5(2) = 22

S_{3} = 3(3)^{2} + 5(3) = 42

S_{4} = 3(4)^{2} + 5(4) = 68

∴ a_{1} = S_{1} = 8

a_{2} = S_{2} – S_{1} ⇒ 22 – 8 ⇒ 14

a_{3} = S_{2} – S_{1} ⇒ 22 – 8 ⇒ 14

a_{4} = S_{2} – S_{1} ⇒ 22 – 8 ⇒ 14

Thus A.P. will be 8, 14, 20, 26 …… 164

a = 8,

d = 14 – 8 = 6 and a_{n} = 164

∴ 164 = a + (n – 1)d

164 = 8 + (n – 1)

(n – 1) = 156/6 = 26

∴ n = 26 + 1 = 27

Hence, option (C) is correct.

Question 6.

If sum of n (RBSESolutions.com) terms of A.P. is S_{n} and S_{2n} = 3S_{n}, then S_{3n} : S_{n} will be :

(A) 10

(B) 11

(C) 6

(D) 4

Solution :

S_{2n} = 3 S_{n}

\(\frac { 2n }{ 2 }\) [2a + (2n – 1)d] = \(\frac { 3n }{ 2 }\) [2a + (n – 1)d]

⇒ 4a + 4nd – 2d = 6a + 3nd – 3d

⇒ nd + d = 2a

Now, S_{3n} : S_{n}

Hence, option (C) is correct.

Question 7.

The first and last terms of an A.P. are 1 and 11 respectively. (RBSESolutions.com) If sum of its terms is 36 then number of terms will be :

(A) 5

(B) 6

(C) 9

(D) 11

Solution :

a= 1, l = 11, S_{n} = 36

S_{n} = \(\frac { n }{ 2 }\) (a + l)

36 = \(\frac { n }{ 2 }\) (1 + 11)

⇒ 36 = \(\frac { n }{ 2 }\) × 12

⇒ 36 = 6n

⇒ n = 6

Hence, option. (B) is correct.

Question 8.

Write 5^{th} term of A.P. 3, 5, 7, 9, …., 201 from (RBSESolutions.com) last.

Solution :

Given A.P.

3, 5, 7, 9 …… 201

First term (a) = 3

Common difference (d) = 5 – 3 = 2

Last term (a_{n}) = 201

Formula : r^{th} term from last = a_{n} – (r – 1)d

5^{th} term from last = 201 – (5 – 1)^{2}

= 201 – 4 × 2

= 201 – 8

= 193

Hence, 5^{th} term from last 193.

Question 9.

If three consecutive (RBSESolutions.com) terms of A.P. are \(\frac { 4 }{ 5 }\), a, 2, then find.

Solution :

Given A.P. \(\frac { 4 }{ 5 }\), a, 2

First term = \(\frac { 4 }{ 5 }\)

Second term a = \(\frac { 4 }{ 5 }\) + d

Third term 2 = \(\frac { 4 }{ 5 }\) + 2d

⇒ 2 – \(\frac { 4 }{ 5 }\) = 2d

⇒ \(\frac { 10-4 }{ 5 }\) = 2d

⇒ \(\frac { 6 }{ 5 }\) = 2d

⇒ 2d = \(\frac { 6 }{ 5 }\)

⇒ d = \(\frac { 3 }{ 5 }\)

Second term a = \(\frac { 4 }{ 5 }\) + d

= \(\frac { 4 }{ 5 }\) + \(\frac { 3 }{ 5 }\) = \(\frac { 7 }{ 5 }\)

Thus, a = \(\frac { 7 }{ 5 }\)

Question 10.

Find the sum of (RBSESolutions.com) first 1000 positive integers.

Solution :

Given A.P.

1, 2, 3, 4, 5 …….. 1000

First term (a) = 1

Common difference (d) = 2 – 1 = 1

Last term (l) = 1000

Formula, S_{n} = \(\frac { n }{ 2 }\) (a + l)

S_{1000} = \(\frac { 1000 }{ 2 }\) (1 + 1000)

= 500 × 1001 = 500500

Hence, sum of first 1000 positive integers.

= 500500

Question 11.

Is any term of (RBSESolutions.com) sequence 5, 11, 17, 23, …. will be 299 ?

Solution :

Given A.P.

5, 11, 17, 23, ……..

First term (a) = 5

Common difference (d) = 11 – 5 = 6

n^{th} term (a_{n}) = 299

Formula a_{n} = a + (n – 1)d

⇒ 299 = 5 + (n – 1) × 6

⇒ 299 – 5 = (n – 1) 6

⇒ 294 = (n – 1) 6

⇒ n – 1 = \(\frac { 294 }{ 6 }\)

⇒ n – 1 = 49

⇒ n = 49 + 1

⇒ n = 50

n is whole no.

∴ Hence, 299 is 50^{th} term of sequence 5, 11, 17, 23 ….

Question 12.

Which term (RBSESolutions.com) of AP. 20, 19 \(\frac { 1 }{ 4 }\), 18 \(\frac { 1 }{ 2 }\), 17 \(\frac { 3 }{ 4 }\),… is first negative term.

Solution :

Given series is A.P. whose

First term (a) = 20 and common difference

(d) = – 3/4

∴ Let the n^{th} term of the given AP the first negative term then

a_{n} < 0

⇒ a + (n – 1)d < O

⇒ 20 + (n – 1) × – \(\frac { 3 }{ 4 }\) < 0

⇒ \(\frac { 83 }{ 4 }\) – \(\frac { 3n }{ 4 }\) < 0

⇒ 83 – 3n < 0 ⇒ 3n > 83

⇒ n > 27 \(\frac { 2 }{ 3 }\) ⇒ n ≥ 28

Thus, 29^{th} term of the given sequence the first negative term.

Question 13.

Four numbers are in A.P. If their (RBSESolutions.com) sum is 20 and sum of their squares is 120, then find the numbers.

Solution :

Let in four numbers,

First number = a – 3d

Second number = a – d

Third number = a + d

Fourth number = a + 3d

Sum of four number is 20

∴ 20 = (a – 3d) + (a – d) + (a + d) + (a + 3d)

⇒ 20 = 4a

⇒ a = 5

Sum of squares of four (RBSESolutions.com) numbers is 120.

∴ (a – 3d)^{2} + (a – d)^{2} + (a + d)^{2} + (a + 3d)^{2} = 120

⇒ [a^{2} + 9d^{2} – 6ad + a^{2} + d^{2} – 2ad + a^{2} + d^{2} + 2ad + a^{2} + 9d^{2} + 6ad] = 120

⇒ 4 (a^{2} + 5d^{2}) = 120

⇒ a^{2} + 5d^{2} = 30 [∵ a = 5]

⇒ 5^{2} + 5d^{2} = 30

⇒ 25 + 5d^{2} = 30

⇒ 5d^{2} = 30 – 25

⇒ 5d^{2} = 5

⇒ d^{2} = 1

⇒ d = ±1

Thus, a = 5 and d = ± 1

∴ Four numbers are 2, 4, 6, 8 or 8, 6, 4, 2

Question 14.

If sum of n terms (RBSESolutions.com) of A.P. is \(\frac { { 3n }^{ 2 } }{ 2 } +\frac { 5n }{ 2 }\) then find its 25^{th} term.

Solution :

Given that, sum of n terms of A.P.

S_{n} = \(\frac { { 3n }^{ 2 } }{ 2 } +\frac { 5n }{ 2 }\)

Hence 25^{th} term will be 76.

Question 15.

The houses of a row are numbered (RBSESolutions.com) consecutively from 1 to 49. Show that there is value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the houses following it. Find the value of x.

Solution :

Consecutively marked number of houses are 1, 2, 3, 4 …… 47, 48, 49.

x is such number so that sum of one side of x numbers = sum of other side number of x d in all the terms of series = 1.

then sum of number from 1 to x – 1, (a) = 12, n = x – 1.

and sum of numbers from x + 1 to 49

Then according (RBSESolutions.com) to question,

Hence, value of x is 35.

We hope the given RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Miscellaneous Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Miscellaneous Exercise, drop a comment below and we will get back to you at the earliest.

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