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 rn 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 → ∞ , rn → 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,
∴ 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
∴ 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 − r) ......(2)
From (1) and (2), we get.
3.3 (1 – r) (1 + r) = 8
or, 1 - r2 = 8/9
or, r2 = 1/9
or,
from (2)
116 videos|164 docs|98 tests
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1. What is an infinite geometric progression? |
2. How can I find the common ratio in an infinite geometric progression? |
3. Can an infinite geometric progression have a common ratio greater than 1? |
4. Is it possible for an infinite geometric progression to have a common ratio less than 0? |
5. What is the formula to find the nth term of an infinite geometric progression? |
116 videos|164 docs|98 tests
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