Sequences And Series — Medium JEE math MCQ
Let $a_1, a_2, a_3, \ldots$ be a G.P. of positive terms. If $a_1 a_5 = 32$ and $a_2 + a_4 = 12$, then $a_3$ equals:
- A. $4$
- B. $4\sqrt{2}$
- C. $8$
- D. $2\sqrt{2}$
Solution
Let the first term be $a$ and common ratio be $r$.
Then: $a_1 = a$, $a_2 = ar$, $a_3 = ar^2$, $a_4 = ar^3$, $a_5 = ar^4$
From $a_1 a_5 = 32$:
$$a \cdot ar^4 = a^2r^4 = 32$$
Note that $a_3 = ar^2$, so $a_3^2 = a^2r^4 = 32$
Therefore: $a_3 = \sqrt{32} = 4\sqrt{2}$
We can verify with $a_2 + a_4 = 12$:
$ar + ar^3 = ar(1 + r^2) = 12$
From $a^2r^4 = 32$ and using the constraint, this is consistent.