Sequence and Series – Geometric Progression – JEE Main 21 Jan 2026 Shift 1

Solution:

Let the common ratio of the G.P. be $r$.
Given: $a_2 a_3 a_4 = 64$
$$(ar)(ar^2)(ar^3) = 64$$
$$a^3 r^6 = 64$$
$$(ar^2)^3 = (4)^3 \Rightarrow ar^2 = 4$$
So, the third term $a_3 = 4$.

Also given: $a_1 + a_3 + a_5 = \frac{813}{7}$
$$a + ar^2 + ar^4 = \frac{813}{7}$$
$$a(1 + r^2 + r^4) = \frac{813}{7}$$
Substituting $a = \frac{4}{r^2}$ (from $ar^2=4$):
$$\frac{4}{r^2}(1 + r^2 + r^4) = \frac{813}{7}$$
$$4\left(\frac{1}{r^2} + 1 + r^2\right) = \frac{813}{7}$$
$$\frac{1}{r^2} + r^2 = \frac{813}{28} – 1 = \frac{785}{28}$$

Let $r^2 = t$. Then $t + \frac{1}{t} = \frac{28^2 + 1}{28} = 28 + \frac{1}{28}$.
Since the G.P. is increasing, $r > 1$, so $r^2 = 28$.

We need to find $S = a_3 + a_5 + a_7$.
$$S = ar^2 + ar^4 + ar^6$$
$$S = ar^2(1 + r^2 + r^4)$$
Substituting $ar^2 = 4$ and $r^2 = 28$:
$$S = 4(1 + 28 + 28^2)$$
$$S = 4(1 + 28 + 784)$$
$$S = 4(813) = 3252$$

Ans. (2)