SSC CHSL 8 Jan 2017 Shift 3 Paper

Three numbers a,b,c are in proportion iff ${\mathit{b}}^2=\mathit{a}\mathit{c}$ where b is the mean proportion and c is the third proportion Mean proportion of two numbers x and y = 16
$\Rightarrow \mathit{x}\mathit{y}={\left(16\right)}^2=256$ -----(i)
Third proportion = 128
⇒ ${\mathit{y}}^2=\mathit{x}\times 128$ ------(ii)
Substituting value of from equation(i) in equation(ii), we get :
⇒ ${\mathit{y}}^2=\frac{256}{\mathit{y}}\times 128$
⇒ ${\mathit{y}}^3={\left(2\right)}^8\times {\left(2\right)}^7={\left(2\right)}^{8+7}$
⇒$\mathit{y}={\left(2\right)}^{\frac{15}{3}}=2^5=32$
Substituting it in equation(i), ⇒ $\mathit{x}=\frac{256}{32}=8$