HOW TO FIND THE SUM OF GEOMETRIC SERIES

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.

s_n  =  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)]

s_n  =  a(rⁿ - 1)/(r - 1),  if r > 1

s_n  =  na,  if r  =  1

  =  (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)

s_n  =  a(rⁿ - 1)/(r - 1),  if r > 1

s_n  =  na,  if r  =  1

  =  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)

s_n  =  a(rⁿ - 1)/(r - 1),  if r > 1

s_n  =  na,  if r  =  1

Hence the sum of the given series is 

s_n  =  n - (2/3)(1 - 0.1ⁿ)

Example 5 :

Sum of the series 1 + 3 + 9 + 27 + .......... is 364. The number of terms is 

Solution :

1 + 3 + 9 + 27 + .......... is 364

The sum of the series = 364

a = 1, r = 3/1 = 3 > 1

s_n = a(rⁿ - 1)/(r - 1)

 1(3ⁿ - 1)/(3 - 1) = 364

 1(3ⁿ - 1)/2 = 364

3ⁿ - 1 = 364(2)

3ⁿ - 1 = 728

Adding 1, we get

3ⁿ  = 728 + 1

3ⁿ  = 729

3ⁿ  = 3⁶

By equating powers, we get

n = 6

So, the sum of 6 terms if the geometric progression is 324.

Example 6 :

The sum of the series 1 + 2 + 4 + 8 +........... to n term

Solution :

1 + 2 + 4 + 8 +........... to n term

a = 1, r = 2/1 = 2 > 1

s_n = a(rⁿ - 1) / (r - 1)

s_n = 1(2ⁿ - 1) / (2 - 1)

s_n = 2ⁿ - 1

So, the sum of the series is 2ⁿ - 1.

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