# GEOMETRIC SERIES

A gematric series is the one which contains the sum of the terms of a geometric sequence.

Geometric Sequence :

a, ar, ar2, ar3, ..........

Geometric Series :

a + ar + ar+ ar+ ..........

## Sum of First n Terms of Geometric Series

To find the sum of n terms of a geometric series, we use one of the formulas given below.

Sn = a1(rn - 1)/(r - 1),    if r ≠ 1

Sn = na1,    if r = 1

where 'a' is the first term of the series and 'r' is the common ratio.

## Sum of Infinite Geometric Series

Formula to find the sum of infinite geometric series :

S = a1/(1 - r),    if -1 < r < 1

where 'a' is the first term of the series and 'r' is the common ratio.

r = second term/first term or 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.

Example 1 :

Find the sum of the following geometric series.

1 + 3 + 9 + .......... to 10 terms

Solution :

a1 = 1

r = a2/a1 = 3/1 = 3 > 1

n = 10

Formula to find sum of n terms of a geometric series where r  1.

Sn = a1(rn - 1)/(r - 1)

Substitute a1 = 1 and r = 3.

S10 = 1(310 - 1)/(3 - 1)

= (59049 - 1)/2

= 59048/2

= 29524

Example 2 :

Find the sum of 8 terms of the following geometric series.

1 + (-3) + 9 + (-27) +..........

Solution :

a1 = 1

r = a2/a1 = -3/1 = -3 < 1

n = 10

Formula to find sum of n terms of a geometric series where r > 1.

Sn = a1(rn - 1)/(r - 1)

Substitute a1 = 1, r = -3 and n = 8.

S8 = 1[(-3)8 - 1]/(-3 - 1)

= (6561 - 1)/(-4)

= 6560/(-4)

= -1640

Example 3 :

Find the sum of 30 terms of the following geometric series.

7 + 7 + 7 + 7 +..........

Solution :

a1 = 7

r = a2/a1 = 7/7 = 1

n = 30

Formula to find sum of n terms of a geometric series where r = 1.

Sn = na1

Substitute a1 = 7 and r = 1.

S30 = 30(7)

= 210

Example 4 :

Find the sum of terms of the following geometric series.

1 + 2 + 4 + 8 +.......... + 2048

Solution :

a1 = 1

r = a2/a1 = 2/1 = 2 > 1

The value of n is not given. So, we have to find the value of n.

Let 2048 be the nth term.

an = 2048

a1rn-1 = 2048

Substitute a1 = 1 and r = 2.

1(2)n-1 = 2048

2n-1 = 211

n - 1 = 11

n = 12

Formula to find sum of n terms of a geometric series where r  1.

Sn = a1(rn - 1)/(r - 1)

Substitute a1 = 1, r = -3 and n = 12.

S12 = 1[212 - 1]/(2 - 1)

= (4096 - 1)/1

= 4095

Example 5 :

Find the sum of the following infinite geometric series.

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.

The sum for the given infinite geometric series exists.

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 6 :

Find the sum of the following infinite geometric series.

-3 + (-12) + (-48) +..........

Solution :

a1 = 3

r = a2/a1 = -12/(-3) = 4

The value of r (= 4) is not in the interval -1 < r < 1.

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

Example 7 :

A ball is dropped from a height of 6 m and on each bounce it bounces 2/3 of its previous height.

(i) What is the total length of the downward paths ?

(ii) What is the total length of the upward paths ?

(iii) How far does the ball travel till it stops bouncing?

Solution :

(i) Distance covered in the downward path :

= 6 + 4 + 8/3 + 16/9 + .............

a1 = 6

r = a2/a1 = 4/6 = 2/3

The value of r (= 2/3) is in the interval -1 < r < 1.

Formula to find sum of infinite geometric series :

S = a1/(1 - r)

Substitute a1 = 6 and r = 2/3.

= 6/(1 - 2/3)

= 6/(1/3)

= 6(3/1)

= 18 m

(ii)  Distance covered in the upward path :

= 4 + 8/3 + 16/9 + ............

S = a1/(1 - r)

Substitute a1 = 4 and r = 2/3.

S = 4/(1 - 2/3)

= 4/(1/3)

= 4(3/1)

= 12 m

(iii) Total distance covered :

= 18 + 12

= 30 m

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