FIND THE VALUE OF AN INFINITE GEOMETRIC SERIES

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Finding the value of an infinite geometric series is to find the sum of the terms in the series.

Formula to find the sum of infinite geometric series :

where -1 < r < 1

In the formula above, a1 is the first term of the series and is the common ratio.

r = second term/first term

or

r =a2/a1

Note :

In an infinite geometric series, if the value of r is not in the interval -1 < r < 1, then the sum does not exist.

Find the values of the following infinite geometric series :

Example 1 :

1 + 3/4 + 9/16 + 27/64 ............

Solution :

In the given geometric series,

a1 = 1

r = a2/a1

= (3/4)/1

= 3/4

Formula to find the sum of an infinite geometric series :

S = a1/(1 - r)

Substitute a1 = 1 and r = 3/4.

S= 1/(1 - 3/4)

= 1/(1/4)

= 1(4/1)

= 4

The value of the given infinite geometric series is 4.

Example 2 :

1 + 2/3 + 4/9 + 8/27 ............

Solution :

In the given geometric series,

a1= 1

r = a2/a1

= (2/3)/1

= 2/3

Formula to find the sum of an infinite geometric series :

S= a1/(1 - r)

Substitute a1= 1 and r = 2/3.

S= 1/(1 - 2/3)

= 1/(1/3)

= 1(3/1)

= 3

The value of the given infinite geometric series is 3.

Example 3 :

1 + 1/2 + 1/4 + 1/8 ............

Solution :

In the given geometric series,

a1= 1

r = a2/a1

= (1/2)/1

= 1/2

Formula to find the sum of an infinite geometric series :

S= a1/(1 - r)

Substitute a1= 1 and r = 1/2.

S= 1/(1 - 1/2)

= 1/(1/2)

= 1(2/1)

= 2

The value of the given infinite geometric series is 2.

Example 4 :

1 + 3/5 + 9/25 + 27/125 ............

Solution :

In the given geometric series,

a1= 1

r = a2/a1

= (3/5)/1

= 3/5

Formula to find the sum of an infinite geometric series :

S= a1/(1 - r)

Substitute a1= 1 and r = 3/5.

S= 1/(1 - 3/5)

= 1/(2/5)

= 1(5/2)

The value of the given infinite geometric series is 5/2.

Example 5 :

1 + 1/4 + 1/16 + 1/64 ............

Solution :

In the given geometric series,

a1= 1

r = a2/a1

= (1/4)/1

= 1/4

Formula to find the sum of an infinite geometric series :

S= a1/(1 - r)

Substitute a1= 1 and r = 1/4.

S= 1/(1 - 1/4)

= 1/(3/4)

= 1(4/3)

The value of the given infinite geometric series is 4/3.

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