JEE Main 2025 — Sequences And Series Question with Solution

We start with the given sequence $a_1,a_2,a_3,\dots$ of an increasing geometric progression (G.P.). Two key conditions are provided:

$a_3\cdot a_5=729$

$a_2+a_4=\frac{111}{4}$

Calculating using given conditions:

Using the sequence terms:

$a_3=a\cdot r^2,a_5=a\cdot r^4,$

Then:

$a_3\cdot a_5=(a\cdot r^2)(a\cdot r^4)=a^2\cdot r^6=729=27^2$

It follows:

$a\cdot r^3=27\Rightarrow a_4=a\cdot r^3=27\text{(i)}$

Using the second condition:

$a_2+a_4=\frac{111}{4}$

Substituting $a_4=27$:

$a_2=\frac{111}{4}-27$

$a_2=a\cdot r=\frac{3}{4}\text{(ii)}$

Solving for $r$ and $a$:

From equation (i) and (ii), we derive $r^2$:

$r^2=\frac{4\cdot 27}{3}=4\times 9=36$

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

Solve for $a$:

Substituting $r=6$ into $a\cdot r=\frac{3}{4}$:

$a\cdot 6=\frac{3}{4}\Rightarrow a=\frac{3}{24}=\frac{1}{8}$

Calculating the requested sum:

The task is to find:

$24(a_1+a_2+a_3)=24(a+ar+a{r}^2)$

Substitute the values:

$=24\times \frac{1}{8}(1+6+6^2)=3\times (1+6+36)=3\times 43=129$

Therefore, the value is 129.