JEE Main 2025 — Sequences And Series Question with Solution

Given the sequence $x_1,x_2,x_3,x_4$ in geometric progression:

$x_1=a$

$x_2=ar$

$x_3=a{r}^2$

$x_4=a{r}^3$

When you subtract 2, 7, 9, and 5 from $x_1,x_2,x_3,x_4$ respectively, the sequence becomes an arithmetic progression. Thus, the new sequence is:

$a-2$

$ar-7$

$a{r}^2-9$

$a{r}^3-5$

For these to form an arithmetic progression, the common differences must be equal, so:

$(ar-7)-(a-2)=(a{r}^2-9)-(ar-7)$

Simplifying gives:

$a(r-1)-5=ar(r-1)-2$

$a(r-1)(r-1)=-3(i)$

$(ar-7)-(a-2)=(a{r}^3-5)-(a{r}^2-9)$

Simplifying gives:

$a(r-1)-5=a{r}^2(r-1)+4$

$a(r-1)(r^2-1)=-9(ii)$

Using the ratio of equations (ii) and (i):

$\frac{a(r-1)(r^2-1)}{a(r-1)(r-1)}=\frac{-9}{-3}$

$r+1=3⟹r=2$

Plugging back into equation (i):

$a(1)(1)=-3⟹a=-3$

So, the sequence $x_1,x_2,x_3,x_4$ is:

$x_1=-3$

$x_2=-6$

$x_3=-12$

$x_4=-24$

The expression for $\frac{1}{24}(x_1\cdot x_2\cdot x_3\cdot x_4)$ is:

$\frac{1}{24}((-3)\cdot (-6)\cdot (-12)\cdot (-24))=216$