JEE Mathematics 2025 Question with Solution

Given: $a_1, a_2, a_3, \ldots$ is a G.P. of increasing positive numbers.

• $a_3 a_5 = 729$
• $a_2 + a_4 = \frac{111}{4}$

Find: $24(a_1 + a_2 + a_3)$

Let the first term be $a_1$ and the common ratio be $r$. Then

$$a_2 = a_1 r, \quad a_3 = a_1 r^2, \quad a_4 = a_1 r^3, \quad a_5 = a_1 r^4$$

Using $a_3 a_5 = 729$,

$$(a_1 r^2)(a_1 r^4) = a_1^2 r^6 = 729$$

So,

$$(a_1 r^3)^2 = 729$$

Since the terms are positive,

$$a_1 r^3 = 27$$

Now use $a_2 + a_4 = \frac{111}{4}$:

$$a_1 r + a_1 r^3 = \frac{111}{4}$$

Substituting $a_1 r^3 = 27$,

$$a_1 r + 27 = \frac{111}{4}$$

Hence,

$$a_1 r = \frac{111}{4} - 27 = \frac{3}{4}$$

Divide the two relations:

$$\frac{a_1 r^3}{a_1 r} = \frac{27}{3/4}$$

Therefore,

$$r^2 = 36$$

Since the G.P. is increasing and positive, $r = 6$. Now,

$$a_1 r = \frac{3}{4} \Rightarrow a_1 = \frac{3/4}{6} = \frac{1}{8}$$

Then

$$a_2 = a_1 r = \frac{1}{8} \cdot 6 = \frac{3}{4}, \quad a_3 = a_1 r^2 = \frac{1}{8} \cdot 36 = \frac{9}{2}$$

So,

$$a_1 + a_2 + a_3 = \frac{1}{8} + \frac{3}{4} + \frac{9}{2} = \frac{43}{8}$$

Therefore,

$$24(a_1 + a_2 + a_3) = 24 \cdot \frac{43}{8} = 129$$

The correct option is C.

Given: $a_3 a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$ for an increasing positive G.P.

Find: $24(a_1 + a_2 + a_3)$

Write

$$a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729$$

So,

$$ar^3 = 27$$

Also,

$$a_2 + a_4 = ar + ar^3 = \frac{111}{4}$$

Hence,

$$ar = \frac{111}{4} - 27 = \frac{3}{4}$$

Now divide:

$$\frac{ar^3}{ar} = r^2 = \frac{27}{3/4} = 36$$

Thus $r = 6$, and

$$a = \frac{ar}{r} = \frac{3/4}{6} = \frac{1}{8}$$

Therefore,

$$24(a_1 + a_2 + a_3) = 24a(1 + r + r^2)$$ $$= 24 \cdot \frac{1}{8}(1 + 6 + 36)$$ $$= 3 \cdot 43 = 129$$

The correct option is C.