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(rn - 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(rn - 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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