Question

If $\frac{\text{(a+b)}}{(1-\mathrm{ab})}\text{, b,}\frac{\text{(b+c)}}{(1-\mathrm{bc})}$ are in AP, then $\text{a,}\frac{\text{1}}{b}\text{, c}$ are in

• A. $\text{A.P}$
• B. $\text{G.P}$
• C. $\text{H.P}$
• D. None of these

Answer 1

The correct option is C

$\text{H.P}$

Explanation for the correct option :

Step $\mathbf{1}$. Given Data

$\frac{\text{(a+b)}}{(1-\mathrm{ab})}\text{, b,}\frac{\text{(b+c)}}{(1-\mathrm{bc})}$ are in Arithmetic Progression (A.P)

Step $\mathbf{2}$. Finding the relation between $\mathbf{\text{a,}}\frac{\mathbf{\text{1}}}{\mathbf{b}}\mathbf{\text{, c}}$

As the above $3$ terms are in A.P

$2\text{b}$ $=$ $\frac{\text{(a+b)}}{(1-\mathrm{ab})}\text{+}\frac{\text{(b+c)}}{(1-\mathrm{bc})}$

$\Rightarrow$ $2\text{b}$ $=$$\frac{\text{(1-bc) (a+b) + (1-ab) (b+c)}}{(1-\mathrm{ab})(1-\mathrm{bc})}$

$\Rightarrow$ $2\text{b}$ $=$ $\frac{{\text{(a – abc + b – b}}^2{\text{c + b – ab}}^2\text{+ c – abc )}}{(1\; –\mathrm{ab}–\mathrm{bc}+{\mathrm{ab}}^2\text{c)}}$

$\Rightarrow$ ${\text{2b (1 – ab – bc + ab}}^2{\text{c) = (a – abc + b – b}}^2{\text{c + b – ab}}^2\text{+ c – abc )}$

$\Rightarrow$ ${\text{2b – 2ab}}^2{\text{– 2b}}^2{\text{c + 2ab}}^3{\text{c = a + 2b + c – 2abc – b}}^2{\text{c – ab}}^2$

$\Rightarrow$ ${\text{-ab}}^2{\text{– b}}^2{\text{c + 2ab}}^3\text{c = a + c – 2abc}$

$\Rightarrow$ ${\text{2ab}}^3{\text{c + 2abc = a + c + ab}}^2{\text{+ b}}^2\text{c}$

$\Rightarrow$ ${\text{2abc( b}}^2{\text{+1) = (a+c) + b}}^2\text{(a+c)}$

$\Rightarrow$ ${\text{2abc( b}}^2{\text{+1) = (a+c) (b}}^2\text{+1)}$

$\Rightarrow$ $\text{2abc = a + c}$

Divide by ac

$\Rightarrow$ $\text{2b =}\frac{1}{c}+\frac{1}{a}$

This implies $\text{a,}\frac{\text{1}}{\mathrm{b}}\text{, c}$ are in H.P.

Hence, option $\left(C\right)$ is correct.