Question

1) Find the sum of the geometric series 3 + 2 + 4 3 + 8 9 + . . .

(2) Prove that the following series converges, and find its sum 0.6 + 0.06 + 0.006 + 0.0006 + 0.00006 + · ·

Answer 1

Solution :-

(a) It should be like this $3+2+\frac{4}{3}+\frac{8}{9}.....$

a=3

r=$\frac{2}{3}$

S_n=$a.\frac{1-r^n}{1-r}$

$S_n=3.\frac{(1-\frac{2}{3}^n)}{(1-\frac{2}{3})}$

put the value of n and get the sum upto n

terms.

As value of

is not given

So we will find sum of in fininite terms

$= \frac{a}{1-r}\\ ={3\over 1-\frac{2}{3}}\\ =9 \ answer$

(b) 0.6 + 0.06 + 0.006 + 0.0006 + 0.00006 + · ·

$S_n=a.\frac{1-r^n}{1-r}$

$S_n={\frac{6}{10}(1-\frac{1}{10}^n)\over( 1-\frac{1}{10})}$ .....(1)

put the value of n and get the sum upto n

terms.

If we simplify the S_n

We got

$S_n=\frac{2}{3}(1-\frac{1}{10^n})$

And as here also n is not given

So we will find sum of infinite terms

$= \frac{a}{1-r}\\ ={\frac{6}{10}\over 1- \frac{1}{10}}\\ =\frac{6}{9} \ answer$

The sum will always come less (using equation 1) than 1 , by the the property of convergence we can say the the series convergence

Hence proved.