Question

If $x^a=x^{b/2}{z}^{b/2}=z^c$, then $a,b,c$ are in:

• A. A.P
• B. G.P
• C. H.P
• D. None of these

Answer 1

The correct option is C

H.P

Explanation for the correct option.

It is given that, $x^a=x^{b/2}{z}^{b/2}$

Taking log on both sides, we get;

$\begin{array}{rcl}a\log x& =& \frac{b}{2}\log x+\frac{b}{2}\log z\\ & \Rightarrow & \frac{b}{2}\log z=\left(a-\frac{b}{2}\right)\log x...\left(1\right)\end{array}$

And it is also given that, $x^a=z^c$

Taking log on both sides, we get;

$\begin{array}{rcl}a\log x& =& c\log z\\ & \Rightarrow & \log z=\frac{a\log x}{c}...\left(2\right)\end{array}$

Substituting the value of $\log z$ in $\left(1\right)$, we get

$\begin{array}{rcl}\frac{b}{2}\times \frac{a}{c}\log x& =& \left(a-\frac{b}{2}\right)\log x\\ \Rightarrow \frac{ab}{2c}& =& a-\frac{b}{2}\end{array}$

Dividing both side by $ab$, we get

$\begin{array}{rcl}\frac{1}{2c}& =& \frac{1}{b}-\frac{1}{2a}\\ & \Rightarrow & \frac{1}{b}=\frac{1}{2}\left(\frac{1}{c}+\frac{1}{a}\right)\\ & \Rightarrow & \frac{2}{b}=\frac{1}{c}+\frac{1}{a}\end{array}$

Therefore, $a,b,c$ are in H.P. as they are satisfying the condition of H.P.

Hence, option C is correct.