FIND THE VALUE OF AN INFINITE GEOMETRIC SERIES

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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