JEE Mathematics 2026 Question with Solution

Given: $a_1, a_2, a_3, \dots$ are in G.P. with increasing positive terms, $a_2 a_3 a_4 = 64$ and $a_1 + a_3 + a_5 = \frac{813}{7}$.

Find: $a_3 + a_5 + a_7$.

Let $a_1 = a$ and common ratio be $r$. Since the terms are increasing and positive, $a>0$ and $r>1$.

Using $a_n = ar^{n-1}$,

$$a_2 a_3 a_4 = (ar)(ar^2)(ar^3) = a^3 r^6 = 64$$

So,

$$(ar^2)^3 = 64 = 4^3$$

Hence,

$$ar^2 = 4$$

Therefore, $a_3 = 4$.

Now use

$$a_1 + a_3 + a_5 = a + ar^2 + ar^4 = \frac{813}{7}$$

Since $ar^2 = 4$, we get

$$a + 4 + 4r^2 = \frac{813}{7}$$

Also, from $ar^2 = 4$,

$$a = \frac{4}{r^2}$$

Substituting,

$$\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$$ $$4\left(\frac{1}{r^2} + 1 + r^2\right) = \frac{813}{7}$$ $$r^2 + \frac{1}{r^2} + 1 = \frac{813}{28}$$ $$r^2 + \frac{1}{r^2} = \frac{785}{28}$$

Let $y = r^2$. Then

$$y + \frac{1}{y} = \frac{785}{28}$$

So,

$$28y^2 - 785y + 28 = 0$$

Solving,

$$y = \frac{785 \pm \sqrt{613089}}{56} = \frac{785 \pm 783}{56}$$

Thus,

$$y = 28 \quad \text{or} \quad y = \frac{1}{28}$$

Since the G.P. is increasing, $r>1$, hence $r^2>1$. Therefore,

$$r^2 = 28$$

Now,

$$a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$$

Factor out $r^2$:

$$a_3 + a_5 + a_7 = r^2(a + ar^2 + ar^4)$$

Using $r^2 = 28$ and $a + ar^2 + ar^4 = \frac{813}{7}$,

$$a_3 + a_5 + a_7 = 28 \times \frac{813}{7} = 3252$$

Therefore, the correct option is D.

Given: $a_2 a_3 a_4 = 64$ and $a_1 + a_3 + a_5 = \frac{813}{7}$.

Find: $a_3 + a_5 + a_7$.

For three consecutive G.P. terms,

$$a_2 a_3 a_4 = a_3^3$$

Hence,

$$a_3^3 = 64 \Rightarrow a_3 = 4$$

Now,

$$a_1 + a_3 + a_5 = a + ar^2 + ar^4 = \frac{813}{7}$$

Using $ar^2 = a_3 = 4$,

$$\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$$

This gives

$$r^2 + \frac{1}{r^2} = \frac{785}{28}$$

So $r^2 = 28$ or $\frac{1}{28}$, and since the terms are increasing, choose

$$r^2 = 28$$

Now observe

$$a_3 + a_5 + a_7 = r^2(a_1 + a_3 + a_5)$$

Therefore,

$$a_3 + a_5 + a_7 = 28 \cdot \frac{813}{7} = 3252$$

Therefore, the correct option is D.