HOW TO FIND THE SUM OF GEOMETRIC SERIES

How to Find the Sum of Geometric Series ?

The series is a geometric series if the terms of the series form a geometric sequence.

To find the sum of n terms of the geometric series, we use one of the formulas given below.

sn  =  a(rn - 1)/(r - 1)    if r > 1

sn  =  a(1 - rn)/(1 - r)  if r < 1

sn  =  a/(1 - r)   if r  =  1

Example 1 :

Find the sum of the following finite series

1 + 0.1 + 0.01 + 0.001 + .......... + (0.1)⁹

Solution :

First term (a)  =  1

r  =  0.1/1 ==>  0.1 < 1

The given geometric series is having 10 terms. So, the value n is 10.

sn  =  a(1 - rn)/(1 - r)  if r < 1

s10  =  1(1 - 0.1n)/(1 - 0.1)

s10  =  (1 - 0.1n)/0.9

Therefore the sum of 10 terms of the geometric series is 

(1 - 0.1n)/0.9

Example 2 :

Find the sum of the following finite series

1 + 11 + 111 + .............. to  20 terms

Solution :

The given series is not geometric series as well arithmetic series.

To convert the given as geometric series, we do the following. 

  =  (9/9) [1 + 11 + 111 + .............. to  20 terms]

  =  (1/9) [9 + 99 + 999 + .............. to  20 terms]

  =  (1/9) [(10-1) + (100-1) + (1000-1) + .............. to  20 terms]

  = 1/9 [(10 + 100 + 1000 +.............. to 20 terms)

 - (1 + 1 + 1 +......to 20 terms)]

a  =  10, r  =  100/10

              r  =  10 > 1

a  =  1     r  =  1

sn  =  a(rn - 1)/(r - 1),  if r > 1

sn  =  na,  if r  =  1

sn  =  10(1020 - 1)/(10 - 1)

sn  =  (10/9)(1020 - 1)

sn  =  20(1)

sn  =  20

  =  (1/9)[ (10/9)(1020 - 1) - 20 ] 

  =  [ (10/81 )(1020 - 1) - (20/9) ] 

Hence the required sum is 

(10/81)(1020 - 1) - (20/9)

 Find the sum of the first n terms of the geometric series

Example 3 :

Find the sum of n terms of the following series

7 + 77 + 777 + ..............

Solution :

  =  [7 + 77 + 777 + ..............to  n terms]

  =  7 [1 + 11 + 111 + ..............to  n terms]

  =  (7/9) [9 + 99 + 999 + .............. to  n terms]

  = (7/9)[(10-1) + (100-1) + (1000-1) + .............. to n terms]

  = 7/9[(10+100+1000+.............. to n terms)

 - (1+1+1+......to n terms)

a  =  10, r  =  100/10

  r  =  10

a  =  1 and r  =  1

sn  =  a(r- 1)/(r - 1),  if r > 1

sn  =  na,  if r  =  1

sn  =  10(10n - 1)/(10 - 1)

sn  =  (10/9)(10n - 1)

sn  =  n(1)

sn  =  n

  =  7/9 [ (10/9)(10n - 1) - n ]

  =  [ (70/81)(10n - 1) - (7n/9) ]

Hence the sum of the given series is 

(70/81)(10n - 1) - (7n/9)

Example 4 :

Find the sum of n terms of the following series

0.4 + 0.94 + 0.994 + ............

Solution :

  =  0.4 + 0.94 + 0.994 + ............

  =  [ (1 - 0.6) + (1 - 0.6) + (1 - 0.06)  + ..............to  n terms]

  =  (1 + 1 + 1 +............to n terms) - (0.6 + 0.06 + 0.006+....... to n terms)

a  =  1 and r  =  1

a  =  0.6 and r  =  0.06/0.6

r  =  6/60

r  =  1/10

r  =  0.1

sn  =  a(r- 1)/(r - 1),  if r > 1

sn  =  na,  if r  =  1

sn  =  n(1)

sn  =  n

sn  =  0.6(1 - 0.1n)/(1 - 0.1)

sn  =  (0.6/0.9)(1 - 0.1n)

sn  =  (2/3)(1 - 0.1n)

Hence the sum of the given series is 

sn  =  n - (2/3)(1 - 0.1n)

After having gone through the stuff given above, we hope that the students would have understood how to find the sum of the geometric series.

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