CMAT 2015 Solved Question Paper

(c) Let the first term be a and common ratio be r, then GP
series will be a, ar, a${\mathrm{r}}^2,{\mathrm{ar}}^3$,...
According to the question,
a${\mathit{r}}^2$ = a + ar
⇒ a${\mathit{r}}^2$ = a(1 + r) ⇒ ${\mathrm{r}}^2$ - r - 1 = 0
⇒ r = $\frac{1\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(-1\right)}}{2\times 1}$
$\left[\because \mathit{x}=\frac{-\mathit{b}\pm \sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}}{2\mathit{a}}\right]$
= $\frac{1\pm \sqrt{1+4}}{2\times 1}$
= $\frac{1\pm \sqrt{5}}{2}$
Since, terms are positive
So, r = ( 1 + $\sqrt{5}$)/2