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

Explanation:
Given: $x,\frac{3}{2},z$ are in $AP$$\therefore 2\times \frac{3}{2}=x+z$$\Rightarrow x+z=3\dots .1$Now, $x,3,z$ are in GP$\therefore 3^2=xz$$\Rightarrow xz=9\dots .2$Let $x,y,z$ are in $HP$$\therefore y=\frac{2xz}{x+z}=\frac{2\times 9}{3}=6$So, $x,6,z$ are in HP.Hence, the correct option is (A).