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If one AM 'A' and two GM $\mathrm{p}$ and $\mathrm{q}$ are inserted between two given numbers, then find the value of $\frac{p^{2}}{q}+\frac{q^{2}}{p}$
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Verified Answer
The correct answer is:
$2 \mathrm{~A}$
Let a and b are two numbers.
Then, $A=\frac{a+b}{2}...(i)$
Also, $a, p, q, b$ are in GP.
$$
\begin{array}{l}
\therefore \frac{p}{a}=\frac{q}{p}=\frac{b}{q} \\
\Rightarrow \frac{p^{2}}{q}=a \text { and } \frac{q^{2}}{p}=b \\
\therefore \frac{p^{2}}{q}+\frac{q^{2}}{p}=a+b=2 A \text { [Using eq. (i)] }
\end{array}
$$
Then, $A=\frac{a+b}{2}...(i)$
Also, $a, p, q, b$ are in GP.
$$
\begin{array}{l}
\therefore \frac{p}{a}=\frac{q}{p}=\frac{b}{q} \\
\Rightarrow \frac{p^{2}}{q}=a \text { and } \frac{q^{2}}{p}=b \\
\therefore \frac{p^{2}}{q}+\frac{q^{2}}{p}=a+b=2 A \text { [Using eq. (i)] }
\end{array}
$$
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