Question

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 GP
• B. $p,q,r$ are in AP
• C. $p,q,r$ are on HP
• D. $p^2,q^2,r^2$ are in AP

Answer 1

The correct option is C $p,q,r$ are on HP

Let $a^p=b^q=c^r=k$
$\therefore a=k^{1/p},b=k^{1/q},c=k^{1/r}$
Since, $a,b,c$ are in GP.,
$\therefore \frac{b}{a}=\frac{c}{b}$

$\frac{k^{1/q}}{k^{1/p}}=\frac{k^{1/r}}{k^{1/q}}$

$k^{1/q-1/p}=k^{1/r-1/q}$

$\frac{1}{q}-\frac{1}{p}=\frac{1}{r}-\frac{1}{q}$

$\therefore \frac{1}{p},\frac{1}{q},\frac{1}{r}$ are in AP.

$\Rightarrow p,q,r$ in HP.