If a be A.M. and p, q be two G.M.'s between two numbers, then 2A … - Vidyadip

MCQ

If a be $\text{A.M.}$ and $p, q$ be two $\text{G.M.'s}$ between two numbers, then $2A$ is equal to:

✓
$\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
B
$\frac{\text{p}^3-\text{q}^3}{\text{pq}}$
C
$\frac{\text{p}^2+\text{q}^2}{2}$
D
$\frac{\text{pq}}{2}.$

✓

Answer

Correct option: A.

$\frac{\text{p}^3+\text{q}^3}{\text{pq}}$

Let the two positive numbers be $a$ and $b.$
$a, A $ and $b$ are in $A.P.$
$\therefore2\text{A}=\text{a}+\text{b}\cdots(\text{i})$
Also, $a, p, q$ and $b$ are in $G.P.$
$\therefore\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}$
Again, $p = ar$ and $q = ar^2........(ii)$
Now, $2A = a + b [$ From $(i)]$
$=\text{a}+\text{a}\Big(\frac{\text{b}}{\text{a}}\Big)$
$=\text{a}+\text{a}\Bigg(\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}\Bigg)^3$
$=\text{a}+\text{ar}^3$
$=\frac{(\text{ar})^2}{\text{ar}^2}+\frac{\big(\text{ar}^2\big)^2}{\text{ar}}$
$=\frac{\text{p}^2}{\text{q}}+\frac{\text{q}^2}{\text{p}} [$Using $(ii)]$
$=\frac{\text{p}^3+\text{q}^3}{\text{pq}}$

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