The sum of the infinite series 1/5 - 2/5^2 + 2^2/5^3 - 2^3/5^4 + ...... is

Question

hartnn · Answer

the common ratio is -2/5, isn't it ?

anonymous · Answer

Be smart

anonymous · Answer

it is a geometric series, use the well known formula for it's sum

hartnn · Answer

First term = 1/5
so sum to infinity = (1/5)/(1-(-2/5))=(1/5)/(7/5)=1/7.

anonymous · Answer

What is the common ratio if it is GP?

anonymous · Answer

@hartnn said it already

anonymous · Answer

nope

anonymous · Answer

the common ratio is -2/5, isn't it ?
that what first answer says.

hartnn · Answer

Actually it is -2/5...why it is not ?? @Zekarias

anonymous · Answer

$$(\frac{-2}{5})^{0}=1$$

anonymous · Answer

Ok if the common ratio is that and the first term is 1/5 write me 1/5 in terms of -2/5

anonymous · Answer

read what i just wrote

anonymous · Answer

But the first term is 1/5 not 1

anonymous · Answer

1/5*(-2/5)^0=1/5

hartnn · Answer

WHY do u want to do that ? there is no reason to express 1/5 in terms of -2/5 to get the sum....the sum to infinity is still 1/7.....

anonymous · Answer

=>1/5=2^0/5^1

anonymous · Answer

correct all of you.
Thanks

anonymous · Answer

the general term of this series is:
$$a*q ^{n}$$
where a=1/5 , q=-2/5 and n goes fomr n=0 to infinity