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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,
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The correct answer is:
\(p, q, r\) are in H.P.
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 $\mathrm{HP}$
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 $\mathrm{HP}$
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