HOW TO FIND THE SUM OF N TERMS IN A GEOMETRIC SERIES

How to Find the Sum of n Terms in a  Geometric Series ?

A series whose terms are in Geometric progression is called Geometric series.

Let a, ar, ar2 , ...arn-1 , ... be the Geometric Progression.

When r ≠ 1

Sn  =  a(rn - 1)/(r - 1)

When r = 1

Sn  =  na

Infinite series :

Sn  =  a/(1 - r)

Question 1 :

Find the sum of first n terms of the G.P

(i) 5, -3, 9/5, -27/25, ...............

Solution :

a = 5, r = t2/t =  -3/5  ≠ 1

Sn  =   a(rn - 1)/(r - 1)

   =   5((-3/5)n - 1)/((-3/5) - 1)

   =   5((-3/5)n - 1)/((-3-5)/5)

   =   5((-3/5)n - 1)/(-8/5)

   =   (-25/8)((-3/5)n - 1)

Sn  =   (25/8)(1  - (-3/5)n)

(ii) 256, 64, 16,…

Solution :

a = 256, r = t2/t =  64/256  =  1/4 ≠ 1

Sn  =   a(rn - 1)/(r - 1)

   =   256((1/4)n - 1)/((1/4) - 1)

   =   256((1/4)n - 1)/(-3/4)

   =   256(-4/3)((1/4)n - 1)

Sn  =  (-1024/3)((1/4)n - 1)

Sn  =  (1024/3)(1 - (1/4)n)

Question 2 :

Find the sum of first six terms of the G.P. 5, 15, 45, …

Solution :

n = 6, a = 5, r = 15/5  =  3  ≠ 1

Sn  =   a(rn - 1)/(r - 1)

   =   5(36 - 1)/(3 - 1)

=  5(729 - 1)/2

=  5(728/2)

=  1820

Question 3 :

Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.

Solution :

S6  =  46872

Sn  =   a(rn - 1)/(r - 1)

r = 5 and n = 6

S6  =   a(56 - 1)/(5 - 1)

46872(4)  =  a(56 - 1)

46872(4)/15624  =  a

a  =  12

Hence the first term is 12.

Finding sum of infinite geometric series

Question 4 :

Find the sum to infinity of

(i) 9 + 3 + 1 + .....

Solution :

Sum of infinite series :

S  =  a/(1-r)

a = 9, r = 3/9  =  1/3

S   =  9/(1-(1/3))

=  9/(2/3)

=  27/2

Hence the required sum is 27/2.

(ii) 21 + 14 + 28/3 + .............

Solution :

a = 21, r = 14/21  =  2/3

Sum of infinite series :

S  =  a/(1-r)

S   =  21/(1-(2/3))

=  21/(1/3)

=  63

Hence the required sum is 63.

After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Sum of n Terms in a Geometric Series". 

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