Question

If $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are in HP, then

• A. $a^{2}b,c^{2}a,b^{2}c$ are in AP
• B. $a^{2}b,b^{2}c,c^{2}a$ are in HP
• C. $a^{2}b,b^{2}c,c^{2}a$ are in GP
• D. None of these

Answer 1

The correct option is A

$a^{2}b,c^{2}a,b^{2}c$ are in AP

Explanation for the correct option:

Step 1. $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are in HP, then the reciprocal will be in AP,

That is, $\frac{b}{a},\frac{c}{b},\frac{a}{c}$ are in AP.

Now,

Common difference will be, $\frac{c}{b}–\frac{b}{a}=\frac{a}{c}–\frac{c}{b}$

$\Rightarrow$ $\frac{(ac-b^2)}{ab}=\frac{(ab-c^2)}{bc}$

Step 2. Eliminate b from both sides and simplify:

$c(ac-b^2)=a(ab-c^2)$

$\Rightarrow$$c^{2}a–b^{2}c=a^{2}b–c^{2}a$

$\Rightarrow$ $2c^{2}a=a^{2}b+b^{2}c$

$\mathbf{\therefore }{\mathit{a}}^{\mathbf{2}}\mathit{b}\mathbf{,}\mathbf{}{\mathit{c}}^{\mathbf{2}}\mathit{a}\mathbf{,}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathit{c}$ are in AP.

Hence, Option ‘A’ is correct.