An infinite geometric series has sum 2014. If the sum of their squares is also 2014, find the first term?

Question

First and last term of a geometric progression are 3 and 96. If the sum of all these terms is 189, then find the number of terms in this progression.

Answer 1

4028/2015

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Si=a/(1-r) - the formula for sum of infinite geometric series, where- a=first term, r=common ratio
a/(1-r)=2014 --------> (1)
a^2/(1-r^2)=2014 --------> (2)
(1)/(2) => a(1-r)/2014/(a^2/(1-r^2))=2014/2014 => a/(1+r)=1 => a=1+r => r=a-1 --------> (3)
if we put (3) into (1), then
a/(1-(a-1))=2014 =>a/(2-a)=2014 => a=2014*(2-a) => 2015a=4028 => a=4028/2015