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**Infinite Geometric Progression **

So far, we have found the sum of a finite number of terms of a G. P. We will now learn to find out the sum of infinitely many terms of a G P such as.

1, 1/2, 1/4, 1/8, 1/16, ......

We will proceed as follows: Here a = 1, r = 1/2

The n^{ th} term of the G. P. is , and sum to n terms

i.e;

So, no matter, how large n may be, the sum of n terms is never more than 2.

So, if we take the sum of all the infinitely many terms, we shall not get more than 2 as answer.

Also note that the recurring decimal 0.3 is really 0.3 + 0.03 + 0.003 + 0.0003 + ... i.e., 0.3 is actually the sum of the above infinite sequence.

On the other hand it is at once obvious that if we sum infinitely many terms of the G. P. 1, 2, 4, 8, 16, ... we shall get a finite sum.

So, sometimes we may be able to add the infinitely many terms of G. P. and sometimes are may not. We shall discuss this question now.

**Sum of Infinite Geometric Progression**

Let us consider a G. P. with infinite number of terms and common ratio r.

**Case 1 **: We assume that | r | > 1

The expression for the sum of n terms of the G. P. is then given by

......(A)

Now as n becomes larger and larger r^{n} also becomes larger and larger. Thus, when n is infinitely large and | r | > 1 then the sum is also infinitely large which has no importance in Mathematics. We now consider the other possibility.

**Case 2 :** Let | r | < 1

Formula (A) can be written as

Now as n becomes infinitely large, rn becomes infinitely small, i.e., as n â†’ âˆž , r^{n} â†’ 0, then the above expression for sum takes the form

Hence, the sum of an infinite G. P. with the first term 'a' and common ratio 'r' is given by

**Example 1. Find the sum of the infinite **

**Solution : **Here, the first term of the infinite G. P. is a = 1/3,

and

Here,

âˆ´ Using the formula for sum we have

Hence, the sum of the given G. P. is 1/5

**Example ****2. Express the recurring decimal **** as an infinite G. P. and find its value inrational form.**

**Solution : ** = 0.3333333.....

= 0.3 + 0.03 + 0.003 + 0.0003 + ....

The above is an infinite G. P. with the first term a = 3/10

and

Hence, by using the formula , we get

Hence, the recurring decimal = 1/3

**Example 3. The distance travelled (in cm) by a simple pendulum in consecutive secondsare 16, 12, 9, ... How much distance will it travel before coming to rest ?**

**Solution : **The distance travelled by the pendulum in consecutive seconds are, 16, 12, 9, ... is an infinite geometric progression with the first term a = 16

and r = 12/16 = 3/4 < 1

Hence, using the formula ,we get

âˆ´ Distance travelled by the pendulum is 64 cm

**Example 4. The sum of an infinite G. P. is 3 and sum of its first two terms is 8/3. Find the first term.**

**Solution : **In this problem S = 3. Let a be the first term and r be the common ratio of the given infinite G. P.

Then according to the question.

a + ar = 8/3

or , 3a(1+r) = 8 ......(1)

Hence, using the formula ,we get

or, a = 3 (1 âˆ’ r) ......(2)

From (1) and (2), we get.

3.3 (1 â€“ r) (1 + r) = 8

or, 1 - r^{2} = 8/9

or, r^{2} = 1/9

or,

from (2)

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