Q152 / 270 · NDA 2026 (Phase 1) · Mathematics
If $p$, $g_1$, $g_2$ and $q$ are in GP and $m$ is the arithmetic mean of $p$ and $q$, then $\dfrac{g_1^2}{g_2}+\dfrac{g_2^2}{g_1}$ is equal to
- A$m$
- Correct answerB$2m$
- C$1$
- D$\dfrac{1}{2}$
Solution
Given $p$, $g_1$, $g_2$, $q$ are in GP.
Let the common ratio of the GP be $r$.
Then $g_1 = pr$, $g_2 = pr^2$, and $q = pr^3$.
The arithmetic mean of $p$ and $q$ is $m$, so $m = \dfrac{p + q}{2} \Rightarrow p + q = 2m$.
Consider the expression $\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1}$.
Substituting the values of $g_1$ and $g_2$:
$\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1} = \dfrac{(pr)^2}{pr^2} + \dfrac{(pr^2)^2}{pr}$
$= \dfrac{p^2r^2}{pr^2} + \dfrac{p^2r^4}{pr}$
$= p + pr^3$
Since $q = pr^3$, we get:
$= p + q$
Substituting $p + q = 2m$:
$\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1} = 2m$
Answer: $2m$
Let the common ratio of the GP be $r$.
Then $g_1 = pr$, $g_2 = pr^2$, and $q = pr^3$.
The arithmetic mean of $p$ and $q$ is $m$, so $m = \dfrac{p + q}{2} \Rightarrow p + q = 2m$.
Consider the expression $\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1}$.
Substituting the values of $g_1$ and $g_2$:
$\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1} = \dfrac{(pr)^2}{pr^2} + \dfrac{(pr^2)^2}{pr}$
$= \dfrac{p^2r^2}{pr^2} + \dfrac{p^2r^4}{pr}$
$= p + pr^3$
Since $q = pr^3$, we get:
$= p + q$
Substituting $p + q = 2m$:
$\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1} = 2m$
Answer: $2m$
Hence, the correct answer is option: 'B'.