# How do you find the value of a if the infinite geometric series #a+ a^2 + 2^3 + ...... =4a#, a cannot equal to 0?

##### 1 Answer

Nov 27, 2015

Assuming the

#(1-a)4a = (1-a)(sum_(n=1)^oo a^n) = a#

Hence

#### Explanation:

#(1-a)(sum_(n=1)^N a^n)#

#=sum_(n=1)^N a^n - a sum_(n=1)^N a^n#

#=a+color(red)(cancel(color(black)(sum_(n=2)^N a^n)))-color(red)(cancel(color(black)(sum_(n=2)^N a^n)))-a^(N+1)#

#=a-a^(N+1)#

If

#(1-a)(sum_(n=1)^oo a^n) = lim_(N->oo) (a-a^(N+1)) = a#

(If

So:

#(1-a)4a = (1-a)(sum_(n=1)^oo a^n) = a#

We are told that

#4(1-a) = 1#

So:

So: