Let a, b, c, p, q and r be positive real numb | vedclass

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(C) Let $a^{p} = b^{q} = c^{r} = k$. Since $a, b, c$ are positive real numbers, we can write $a = k^{1/p}$, $b = k^{1/q}$, and $c = k^{1/r}$. Given th

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$p, q, r$ are in $H.P.$

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(C) Let $a^{p} = b^{q} = c^{r} = k$. Since $a, b, c$ are positive real numbers, we can write $a = k^{1/p}$, $b = k^{1/q}$, and $c = k^{1/r}$. Given that $a, b, c$ are in $GP$, we have $\frac{b}{a} = \frac{c}{b}$. Substituting the values of $a, b, c$, we get $\frac{k^{1/q}}{k^{1/p}} = \frac{k^{1/r}}{k^{1/q}}$. This implies $k^{(1/q - 1/p)} = k^{(1/r - 1/q)}$. Equating the exponents, we get $\frac{1}{q} - \frac{1}{p} = \frac{1}{r} - \frac{1}{q}$, which simplifies to $\frac{2}{q} = \frac{1}{p} + \frac{1}{r}$. This shows that $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in $A.P.$ Therefore, $p, q, r$ are in $H.P.$

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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