# Class 11 Maths MCQ – Arithmetic Progression (A.P.) – 2

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 nth term is 42, then find n.
a) 10
b) 11
c) 12
d) 14

Explanation: We know, a=20, d=2, an=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 nth term is 42, then find sum up to n terms.
a) 12
b) 42
c) 352
d) 372

Explanation: We know, a=20, d=2, an=42.
a+(n-1) d = 42 => 20 + 2(n-1) = 42
=>2(n-1) = 42-20=22 => n-1 = 11 => n=12.
Sn = $$\frac{n}{2}$$ (a+l) => S12 = $$\frac{12}{2}$$ (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

Explanation: Given, an = 3n.
We know, d = an-an-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 8th terms.
a) 4:5
b) 5:4
c) 3:10
d) 3:11

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 8th 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

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.
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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

Explanation: Let A.P. be 2, A1, A2, A3, A4, 22.
=>a=2 and a6 = a+5d = 22 => 2+5*d=22 => d=4.
A1 = a2 = a + d = 2 + 4 = 6.
A2 = A1 + d = 6 + 4 = 10.
A3 = 10 + 4 = 14.
A4 = 14 + 4 = 18.

7. In A.P. 171, 162, 153, ………. Find first negative term.
a) 0
b) -2
c) -6
d) -9

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

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

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.