JEE Main 2026 — 21 January, Morning Shift — Question 6

Let the G.P. be $a, ar, ar^2, ar^3, \ldots$ with $a>0, r>1$. Given $a_2 \cdot a_3 \cdot a_4 = 64$: $(ar)(ar^2)(ar^3) = a^3 r^6 = 64$. Thus $(ar^2)^3 = 64 \Rightarrow ar^2 = 4$. Given $a_1 + a_3 + a_5 = \frac{813}{7}$: $a + ar^2 + ar^4 = \frac{813}{7}$. Substitute $ar^2 = 4$: $a + 4 + ar^4 = \frac{813}{7}$. Since $ar^4 = (ar^2)r^2 = 4r^2$, we have $a + 4 + 4r^2 = \frac{813}{7}$. Also $a = \frac{4}{r^2}$. So $\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$. Multiply by $7r^2$: $28 + 28r^2 + 28r^4 = 813r^2$. Thus $28r^4 - 785r^2 + 28 = 0$. Solve: $r^2 = 28$ (since $r>1$). Now $a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6 = ar^2(1 + r^2 + r^4) = 4(1 + 28 + 784) = 4 \times 813 = 3252$.