# Mathematics Questions and Answers – Arithmetic Progression(A.P.) – 1

«
»

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Arithmetic Progression(A.P.) – 1”.

1. A sequence is called ___________________ if an+1 = an + d.
a) arithmetic progression
b) geometric Progression
c) harmonic Progression
d) special Progression

Explanation: A sequence is called arithmetic progression if an+1 = an + d where a1 is the first term and d is common difference.

2. What is nth term of an A.P.?
a) an = a + (n-1) d
b) an = a + (n) d
c) an = a*rn-1
d) an = a*rn

Explanation: Since every term of an A.P. is incremented by common difference d.
i.e. an+1 = an + d = an-1 + 2d = ……. = a1 + n*d
or an = a + (n-1) d

3. If an A.P. is 3,5,7,9……. Find the 12th term of the A.P.
a) 12
b) 21
c) 22
d) 25

Explanation: From the given A.P., a=3 and d=5-3 =2.
We know, an= a + (n-1) d => a12 = a+11d = 3+11*2 = 3+22 = 25.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. If a constant is added or subtracted from each term of an A.P. then resulting sequence is also an A.P.
a) True
b) False

Explanation: Let x be the constant which is added to each term of an A.P., then an’ = an + x and a’ = a + x. So nth term will be an’ = an + x = a+(n-1) d + x = (a + x) + (n-1) d = a’ + (n-1) d which is nth term of an A.P. If x is negative it is case of subtraction.

5. If a constant is multiplied to A.P. then resulting sequence is also an A.P.
a) True
b) False

Explanation: Let x be the constant which is multiplied to each term of an A.P., then an’ = an * x and a’ = a * x. So nth term will be an’ = an * x = (a+(n-1) d) * x = (a * x) + (n-1) d x = a’ + (n-1) x d which is nth term of an A.P.

6. If 3rd term of an A.P. is 6 and 5th term of that A.P. is 12. Then find the 21st term of that A.P.
a) 40
b) 42
c) 60
d) 63

Explanation: Given, a3 = 6 and a5 = 12.
=> a + 2d = 6 and a + 4d = 12
=> 2d = 6 => d=3.
=> a + 2*3 = 6 => a=0
So, a21 = a + 20 d = 0 + 20*3 = 60.

7. If sum of n terms of an A.P. is n2+5n then find general term.
a) n+1
b) 2n
c) 3n
d) n2+3n

Explanation: Given, Sn = n2+5n
We know, an = Sn– Sn-1 = (n2+5n) – ((n-1)2+5(n-1)) = (n2+5n) – (n2+1-2n+5n-1) = 2n.

8. If an A.P. is 1,7,13, 19, ……… Find the sum of 22 terms.
a) 127
b) 1204
c) 1408
d) 1604

Explanation: From the given A.P., a=1 and d=7-1 = 6.
We know, Sn = $$\frac{n}{2} (2a+(n-1)d)$$
S22 = $$\frac{22}{2} (2*1+(22-1)6)$$ = 11(2+21*6) = 11(2+126) = 11*128 = 1408.

9. If in an A.P., first term is 20 and 12th term is 120. Find the sum up to 12th term.
a) 420
b) 840
c) 140
d) 1680

Explanation: Given, a=20, a12= 120, n=12.
Sn = $$\frac{n}{2}$$ (a+l) => S12 = $$\frac{12}{2} (20+120)$$ = 6*140 = 840.