# SUM OF INFINITE GEOMETRIC 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 r 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.

Examples 1-6 : Find the sum of the following infinite geometric series.

Example 1 :

3 + 1 + 1/3 +..........

Solution :

a1 = 3

r = a2/a1 = 1/3

The value of r (= 1/3) is in the interval -1 < r < 1.

So, the sum for the given infinite geometric series exists.

Formula to find the sum of infinite geometric series :

S = a1/(1 - r)

Substitute a1 = 3 and r = 1/3.

S = 3/(1 - 1/3)

= 3/(2/3)

= 3(3/2)

= 9/2

Example 2 :

-3 - 12 - 48 +..........

Solution :

a1 = 3

r = a2/a1 = -12/(-3) = 4

Since the value of r (= 4) is not in the interval -1 < r < 1, the sum for the given infinite geometric series does not exist.

Example 3 :

1 - 3 - 48 +..........

Solution :

a1 = 1

r = a2/a1 = -3/1 = -3

Since the value of r (= -3) is not in the interval -1 < r < 1, the sum for the given infinite geometric series does not exist.

Example 4 :

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

Solution :

a1 = 1

r = a2/a1 = (1/2)/1 = 1/2

The value of r (= 1/2) is in the interval -1 < r < 1.

So, the sum for the given infinite geometric series exists.

Formula to find the sum of 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

Example 5 :

1 - 0.5 + 0.25 + ..........

Solution :

a1 = 1

r = a2/a1 = -0.5/1 = -0.5

The value of r (= -0.5) is in the interval -1 < r < 1.

So, the sum for the given infinite geometric series exists.

Formula to find the sum of infinite geometric series :

S = a1/(1 - r)

Substitute a1 = 1 and r = -0.5.

S = 1/[1 - (-0.5)]

= 1/(1 + 0.5)

= 1/(1.5)

= 1/(3/2)

= 1(2/3)

= 2/3

Example 6 :

k = 1k =  5(-1/5)k-1

Solution :

Let ak = 5(-1/5)k-1.

When k = 1,

a1 = 5(-1/5)0

= 5(1)

= 5

When k = 2,

a2 = 5(-1/5)2-1

5(-1/5)1

= 5(-1/5)

= -1

When k = 3,

a3 = 5(-1/5)3-1

5(-1/5)2

= 5(1/25)

= 1/5

k = 1k =  5(-1/5)k-1 = 5 - 1 + 1/5 + ..........

a1 = 5

r = a2/a1 = -1/5

The value of r (= -1/5) is in the interval -1 < r < 1.

So, the sum for the given infinite geometric series exists.

Formula to find the sum of infinite geometric series :

S = a1/(1 - r)

Substitute a1 = 5 and r = -1/5.

S = 5/[1 - (-1/5)]

= 5/(1 + 1/5)

= 5/(6/5)

= 5(5/6)

= 25/6

Examples 7-8 : Determine the common ratio of the infinite geometric series.

Example 7 :

a1 = 1 and S = 1.25

Solution :

S = 1.25

a1/(1 - r) = 1.25

1/(1 - r) = 1.25

Take reciprocal on both sides.

(1 - r)/1 = 1/1.25

1 - r = 0.8

Subtract 1 from both sides.

-r = -0.2

Multiply both sides by -1.

r = 0.2

Example 8 :

a1 = -4 and S = -16/5

Solution :

S = -16/5

a1/(1 - r) = -16/5

-4/(1 - r) = -16/5

Take reciprocal on both sides.

(1 - r)/(-4) = -5/16

Multiply both sides by -4.

1 - r = 5/4

Subtract 1 from both sides.

-r = 1/4

Multiply both sides by -1.

r = -1/4

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