Question

If $x,\frac{3}{2},z$ are in AP; $x,3,z$ are in GP; then which one of the following will be in HP?

• A. $x,6,z$
• B. $x,4,z$
• C. $x,2,z$
• D. $x,1,z$

Answer 1

The correct option is A $x,6,z$

If $x,\frac{3}{2},z$ are in A.P

$x,3,z$ in G.P.

$\frac{3}{2}=\left(\frac{x+z}{2}\right)....\left(1\right)$ and $3=\sqrt{xz}\Rightarrow 9=xz...\left(2\right)$

Divide $\left(1\right)$ by $\left(2\right)$ we get

$\frac{6}{\frac{2}{9}}=\frac{x+z}{xz}\to \frac{1}{3}=\frac{1}{z}+\frac{1}{x}\to \frac{2}{6}=\frac{1}{z}+\frac{1}{x}$

$\therefore x,6,z$ are in H.P.