Q117 / 160 · WBJEE 2012 · Mathematics

Let $a, b, c, p, q$ and $r$ be positive real numbers such that $a, b$ and $c$ are in GP and $a^{p}=b^{q}=c^{r}$. Then,

A
$p, q, r$ are in G.P.
B
$p, q, r$ are in A.P.
Correct answer
C
$p, q, r$ are in H.P.
D
$p^{2}, q^{2}, r^{2}$ are in A.P.

Solution

Let $a^{p}=b^{4}=c^{r}=k$ $\therefore \quad a=k^{1 / p}, b=k^{1 / q}, c=k^{1 / r}$ since, $a, b, c$ are in $\mathrm{GP} .$ $\therefore$ $\frac{b}{a}=\frac{c}{b}$ $\frac{k^{1 / q}}{k^{1 / P}}=\frac{k^{1 / r}}{k^{1 / q}}$ $k^{1 / q-1 / p}=k^{1 / r-1 / 4}$ $\frac{1}{q}-\frac{1}{p}=\frac{1}{r}-\frac{1}{q}$ $\therefore \quad \frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in GP. $\Rightarrow \quad p, q, r$ in $\mathrm{HP}$

Hence, the correct answer is option: 'C'.

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Solution

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