## Simple Problems On Geometric Progression

Dear Reader,

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) 1/2^{n} b)2^{n} c) -1/4^{n} d)1/4^{n}

**Answer :** a) 1/2^{n}

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 = 1/2^{n} (since (-1)^{n} = 1 when n is an even number)

Hence, the answer is 1/2^{n}.

**Question 2**

The total number of terms in the series 1, 3, 9,....,19683 is:

a) 11 b) 10 c) 21 c) 17

**Answer:** b) 10

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)} = 3^{9}

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

**Answer:** a) 75C/2

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 - r^{n}) / (1 - r)

Now, S(n) = 45[1 - (-1/5)^{15} ] /(1-(-1/5))

= 45 x [1 + 1/5^{15}] / (6/5)

= 45 x 5/6 x [1 + 1/5^{15}]

= 75/2 x [1 + 1/5^{15}]

= 75C/2 where C = 1 + 1/5^{15} 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(4^{20} - 1)/2 b) 2(4^{20} - 1)/3 c) 2(1-4^{20}) d) 2(1+4^{20})

**Answer:** b) 2(4^{20} - 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(r^{n} - 1) / (r-1)

Now, S(n) = 2(4^{20} - 1) / (4-1)

= 2(4^{20} - 1)/3

Hence, the answer is option b.