Let, first term be a and common ratio is r.
So, [tex]\dfrac{a}{r}+\dfrac{a}{r^2}=96[/tex]
Sum of series, [tex]S = \dfrac{a}{1-r}=500[/tex]
a = 500( 1 - r )
[tex]a( \dfrac{1}{r}+\dfrac{1}{r^2})=96\\\\r=0.916[/tex]
Therefore, the value of common ratio is 0.916 .
Hence, this is the required solution.
