This set of Class 11 Maths Chapter 9 Multiple Choice Questions & Answers (MCQs) focuses on “Arithmetic Progression (A.P.) – 2”.

1. If in an A.P., first term is 20, common difference is 2 and n^{th} term is 42, then find n.

a) 10

b) 11

c) 12

d) 14

View Answer

Explanation: We know, a=20, d=2, a

_{n}=42.

a+(n-1) d = 42 => 20 + 2(n-1) = 42

=>2(n-1) = 42-20=22 => n-1 =11 => n=12.

2. If in an A.P., first term is 20, common difference is 2 and n^{th} term is 42, then find sum up to n terms.

a) 12

b) 42

c) 352

d) 372

View Answer

Explanation: We know, a=20, d=2, a

_{n}=42.

a+(n-1) d = 42 => 20 + 2(n-1) = 42

=>2(n-1) = 42-20=22 => n-1 = 11 => n=12.

S

_{n}= \(\frac{n}{2}\) (a+l) => S

_{12}= \(\frac{12}{2}\) (20+42) = 6*62 = 372.

3. If general term of an A.P. is 3n then find common difference.

a) 2

b) 3

c) 5

d) 6

View Answer

Explanation: Given, a

_{n}= 3n.

We know, d = a

_{n}-a

_{n-1}= 3n – 3(n-1) = 3.

4. The sum of n terms of two arithmetic progressions are in the ratio (2n + 3):(7n + 5). Find the ratio of their 8^{th} terms.

a) 4:5

b) 5:4

c) 3:10

d) 3:11

View Answer

Explanation: Let a, a’ be the first terms and d, d’ be the common differences of 2 A.P.’s respectively.

Given, \(\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a’+(n-1)d’]} = \frac{2n+3}{7n+5}\)

=>\(\frac{a+(n-1)d/2}{a’+(n-1) d’/2} = \frac{2n+3}{7n+5}\)

If we have to find ratio of 8

^{th}terms then (n-1)/2 = 7 => n=15

=>\(\frac{a+7d}{a’+7d’} = \frac{2*15+3}{7*15+5} = \frac{30+3}{105+5} = \frac{33}{110}\) = 3/10.

5. If two numbers are 2 and 6 then find their arithmetic mean.

a) 3

b) 4

c) 5

d) 8

View Answer

Explanation: We know that arithmetic mean of two numbers is given by the average of two numbers i.e. A.M. = (2+6)/2=8/2 = 4.

6. Insert 4 numbers between 2 and 22 such that the resulting sequence is an A.P.

a) 4, 8, 12, 16

b) 5, 9, 13, 17

c) 4, 10, 15, 19

d) 6, 10, 14, 18

View Answer

Explanation: Let A.P. be 2, A

_{1}, A

_{2}, A

_{3}, A

_{4}, 22.

=>a=2 and a

_{6}= a+5d = 22 => 2+5*d=22 => d=4.

A

_{1}= a

_{2}= a + d = 2 + 4 = 6.

A

_{2}= A

_{1}+ d = 6 + 4 = 10.

A

_{3}= 10 + 4 = 14.

A

_{4}= 14 + 4 = 18.

7. In A.P. 171, 162, 153, ………. Find first negative term.

a) 0

b) -2

c) -6

d) -9

View Answer

Explanation: a=171 and d=162-171 = -9.

a

_{n}<0

=>171+(n-1) (-9) < 0

=>180-9n < 0

=>9n > 180

=>n > 20 => n=21 for first negative term.

First negative term is 171+(20) (-9) = 171-180 = -9

8. In A.P. 171, 162, 153, ………. Find first non-positive term.

a) 0

b) -2

c) -6

d) -9

View Answer

Explanation: a=171 and d=162-171 = -9.

a

_{n}<=0

=>171+(n-1) (-9) <=0

=>180-9n <=0

=>9n >= 180

=> n >= 20 => n=20 for first non-positive term.

First negative term is 171+(19) (-9) = 171-171 = 0.

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