Question

The first and second terms of both an $A.P.$ and a $G.P.$ are same $x$ and $y$ respectively where both $x$ and $y$ are $+ive$ and $x$ is greater than $y$. If $S$ be the sum of infinite $G.P.$, then prove that sum of first $n$ terms of $A.P.$ is given by $nx-\frac{n(n-1)x^2}{2S}$.

Answer 1

$d=y-x$ for $A.P.$
$r=\frac{y}{x}<1$ for $G.P$. as $x>y$
$S=\frac{a}{1-r}=\frac{x}{1-y/x}=\frac{x^2}{x-y}$ ......$\left(1\right)$
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{n}{2}[2x+(n-1\left)\right(y-x\left)\right]$
$=nx-\frac{n(n-1)x^2}{2S}$ by $\left(1\right)$