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

Answer:

Assuming the #2^3# should have been #a^3#...

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

Hence #a = 3/4#

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 #abs(a) < 1# then

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

(If #abs(a) >= 1# then the sum does not converge)

So:

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

We are told that #a!=0#, so we can divide both sides by #a# to get:

#4(1-a) = 1#

So: #1-a = 1/4#

So: #a = 1-1/4 = 3/4#