SUM OF INFINITE SERIES OF GEOMETRIC PROGRESSION

Find these sums to infinity, where they exist.

Problem 1 :

80 + 20 + 5 + 1.25 + …

Solution :

Let a₁ = first term and r = common ratio

Here a₁ = 80 and r = a₂/a₁

= 20/80

= 1/4

The value of r = 1/4 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 = 80 and r = 1/4.

S_∞ = 80/(1 – 1/4)

= 80/(3/4)

= 80(4/3)

= 320/3

Problem 2 :

180 – 60 + 20 – 20/3 + …

Solution :

a₁ = 180

 r = a₂/a₁

= -60/180

= -1/3

Since the value of r = -1/3 is not in the interval -1 < r < 1.

S_∞ = 180/(1 – (-1/3))

S_∞ = 180/(1 + (1/3))

S_∞ = 180(3/4)

S_∞ = 135

Problem 3 :

2 + 1.98 + 1.9602 + …

Solution :

a₁ = 2

 r = a₂/a₁

= 1.98/2

= 0.99

The value of r = 0.99 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 = 2 and r = 0.99.

S_∞ = 2/(1 – 0.99)

= 2/(0.01)

= 200

Problem 4 :

-100 + 110 – 121 + …

Solution :

 a₁ = -100

 r = a₂/a₁

= 110/-100

= -11/10

Since the value of r = -11/10 is not in the interval -1 < r < 1.

So, the sum for the given infinite geometric series does not exist.

Problem 5 :

1/10 + 1/100 + 1/1000 + …

Solution :

a₁ = 1/10

 r = a₂/a₁

= (1/100) / (1/10)

= 1/10

The value of r = 1/10 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 = 1/10 and r = 1/10.

S_∞ = (1/10) / (1 – 1/10)

= (1/10) / (9/10)

= 1/9