# Mathematics Questions and Answers – Geometric Progression(G.P.)

«
»

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Geometric Progression(G.P.)”.

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

Explanation: A sequence is called geometric progression if an+1 = an * r where a1 is the first term and r is common ratio.

2. What is nth term of a G.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 G.P. is r times the previous term.
i.e. an+1 = an * r = an-1 * r2 = ….. = a1 * rn
or an = a*rn-1

3. If first term of a G.P. is 20 and common ratio is 4. Find the 5th term.
a) 10240
b) 40960
c) 5120
d) 2560

Explanation: Given, a=20 and r=4.
We know, an = arn-1
=>a5 = 20*44 = 20*256 = 5120.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. If a sequence is in the form 2*5n then which of the following may be the sequence?
a) Arithmetic progression
b) Geometric Progression
c) Harmonic Progression
d) Special Progression

Explanation: If an = 2*5n then
a1 =10, a2 = 50, a3=250.
This is a geometric progression with first term 10 and common ratio 5.

5. If r=1 in a G.P. then what is the sum to n terms?
a) n*a
b) a/n
c) (n-1) a
d) (n+1) a

Explanation: If a is the first term of G.P., then G.P. look like a, a, a, a, …………
Then sum to n terms becomes n*a.

6. If a=3 and r=2 then find the sum up 5th term.
a) 95
b) 82
c) 93
d) 97

Explanation: We know, Sn = a(rn-1)/(r-1)
Here a = 3, r = 2 and n = 5
S5 = 3 (25-1) / (2-1) = 3(32 -1) = 3*31 = 93.

7. In G.P. 4, 8, 16, 32, ………… find the sum up to 5th term.
a) 16
b) 64
c) 128
d) 124

Explanation: In the given G.P., a=4 and r=8/4=2.
We know, Sn = a(rn-1)/(r-1)
=>S5 = 4(25-1) / (2-1) = 4*31 = 124.

8. Which term of G.P. 25, 125, 625, …………. is 390625?
a) 5
b) 6
c) 7
d) 8

Explanation: In the given G.P., a=25 and r = 125/25 = 5.
Given, an = 390625 => arn-1 = 390625
=>25*5n-1 = 390625
=> 5n-1 = 390625/25 = 15625 = 56
=> n-1 = 6 => n=7.

9. In a G.P., 5th term is 27 and 8th term is 729. Find its 11th term.
a) 729
b) 2187
c) 6561
d) 19683

Explanation: Given, a5 = 27 and a8 = 729.
=>ar4 = 27 and ar7 = 729
On dividing we get, r3 = 27 => r=3
=> a=27 / (34) = 1/3
=>a11 = ar10 = (1/3) (310) = 39 = 19683.

10. Find the sum of series 1+1/2 + 1/4 + ………. up to 6 terms.
a) 63/32
b) 32/63
c) 26/53
d) 53/26

Explanation: Given series is G.P. with first term 1 and common ratio 1/2.
We know, Sn = a(1-rn)/(1-r) for r<1.
S6 = 1(1-(1/2)6) / (1-1/2) = (1-1/64) / (1/2) = 63*2/64 = 63/32.

11. How many terms of G.P. 2,4,8,16, …………… are required to give sum 254?
a) 4
b) 5
c) 6
d) 7

Explanation: a=2 and r = 4/2 = 2.
We know, Sn = a(rn-1)/(r-1)
2(2n-1) / (2-1) = 254
=>2n-1 = 127 => 2n = 128 = 27
=> n=7.

12. The sum of first three terms of a G.P. is 21/2 and their product is 27. Find the common ratio.
a) 2
b) 1/2
c) 2 or 1/2
d) neither 2 nor 1/2

Explanation: Let three terms be a/r, a, a*r.
Product = 27 => (a/r) (a) (a*r) = 27 => a3 = 27
=>a = 3.
Sum = 21/2 => (a / r + a + a*r) = 21/2 => a (1 / r + 1 + 1*r) = 21/2
=> (1 / r + 1 + 1*r) = (21/2)/3 = 7/2
=> (r2 + r + 1) = (7/2) r => r2 – (5/2) r + 1 = 0
=> r = 2 and 1/2.

13. The sum of first three terms of a G.P. is 21/2 and their product is 27. Which of the following is not a term of the G.P. if the numbers are positive?
a) 3
b) 2/3
c) 3/2
d) 6

Explanation: Let three terms be a/r, a, a*r.
Product = 27 => (a/r) (a) (a*r) = 27 => a3 = 27
=>a = 3.
Sum = 21/2 => (a / r + a + a*r) = 21/2 => a (1 / r + 1 + 1*r) = 21/2
=> (1 / r + 1 + 1*r) = (21/2)/3 = 7/2
=> (r2 + r + 1) = (7/2) r => r2 – (5/2) r +1 = 0
=> r = 2 and 1/2.
Terms are 3/2, 3, 3*2 i.e. 3/2, 3, 6.

14. Which of the following is the geometric mean of 3 and 12.
a) 4
b) 6
c) 9
d) 10

Explanation: We know, geometric mean of two numbers a and b is given by
G.M. = $$\sqrt{a*b}$$
So, G.M. of 3 and 12 is $$\sqrt{3*12} = \sqrt{36}$$ = 6.

15. If three positive numbers are inserted between 4 and 512 such that the resulting sequence is a G.P., which of the following is not among the numbers inserted?
a) 256
b) 16
c) 64
d) 128

Explanation: Let G.P. be 4, G1, G2, G3, 512.
=>a=4 and a5 = a*r4 = 512 => 4*r4 = 512 => r4 = 512/4 = 128 => r = 4.
G1 = a2 = a * r = 4*4 = 16.
G2 = G1 * r = 16 * 4 = 64.
G3 = G2 * r = 64*4 = 256.

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

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.