Question

If $a,b,c$are in A.P., then$2^{ax+1},2^{bx+1},2^{cx+1},x\ne 0$ are in

• A. $A.P$
• B. $G.P$ only when $x>0$
• C. $G.P$ if $x<0$
• D. $G.P$ for all $x\ne 0$

Answer 1

The correct option is D

$G.P$ for all $x\ne 0$

Solving the series in AP:

Given $a,b,c$are in AP ,

So,

$$\begin{gathered} 2b=a+c \\ {\left(2^{bx+1}\right)}^2=\left(2^{2bx+2}\right) \\ =2^{(a+c)x+2} \\ =2^{ax+1}\times 2^{cx+1} \end{gathered}$$

Therefore,$2^{ax+1},2^{bx+1},2^{cx+1}$ are in G.P

Hence, the terms $2^{ax+1},2^{bx+1},2^{cx+1},x\ne 0$ are in $G.P$so, option (D) is correct.