## Simple Problems On Geometric Progression

Below are four problems on geometric series with simple calculations

Question 1

Find the n-th term of the series -1/2, 1/4, -1/8,...(when n is an even number)
a) 1/2n b)2n c) -1/4n d)1/4n

Answer : a) 1/2n

Solution :

The given series -1/2, 1/4, -1/8,... is a geometric series where a = -1/2
and r = [1/4 / (-1/2)] = -1/2
Then, the nth term of the G.P = tn = a(r(n-1))
tn = (-1/2)x(-1/2)(n-1)
tn = (-1/2)n = (-1)n x (1/2)n
tn = 1/2n (since (-1)n = 1 when n is an even number)

Hence, the answer is 1/2n.

Question 2

The total number of terms in the series 1, 3, 9,....,19683 is:
a) 11 b) 10 c) 21 c) 17

Solution:

The given geometric series is 1, 3, 9,....,19683 where a = 1, r = 3 and last term = 19683
Let us find the total number of terms
i.e, we have to find the position of the last term(19683).
The nth term in G.P represented by tn = ar(n-1)
Now, tn = 1 x 3(n-1) = 19683
1 x 3(n-1) = 19683
3(n-1) = 39
n-1 = 9
n = 10.
Therefore, the nth term of the series is 19683.
Hence, there are 10 terms in the given G.P

Question 3

The sum of the first 15 terms of the series 45 + -9 + 9/5 + -9/25 + .... is equal to:
(here C is a constant)
a) 75C/2 b) 45C/2 c) 45C/4 d) 54C

Solution:

The given G.P series is 45, (-9), 9/5, (-9/25), ....

Here, a = 45, r = -1/5 and n = 15

The sum of the n terms in G.P series when r < 1 = S(n) = a(1 - rn) / (1 - r)

Now, S(n) = 45[1 - (-1/5)15 ] /(1-(-1/5))
= 45 x [1 + 1/515] / (6/5)
= 45 x 5/6 x [1 + 1/515]
= 75/2 x [1 + 1/515]
= 75C/2 where C = 1 + 1/515 is a constant.
Hence the answer is option a.

Question 4

If there are 20 terms in the geometric series 2, 8, 32, 128,.... then the sum of all the terms is equal to:
a) 3(420 - 1)/2 b) 2(420 - 1)/3 c) 2(1-420) d) 2(1+420)
Answer: b) 2(420 - 1)/3

Solution:

The Given G.P series is 2, 8, 32, 128,....
Then we have a = 2, r = 8/2 = 4 and n = 20

The sum of the n terms in G.P series when r > 1 = S(n) = a(rn - 1) / (r-1)

Now, S(n) = 2(420 - 1) / (4-1)
= 2(420 - 1)/3

Hence, the answer is option b.

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