JEE Main 2025 — 7 April, Morning Shift — Question 35

$x_{1}=a ; x_{2}=a r ; x_{3}=a r^{2} ; x_{4}=a r^{3}$

$a-2, a r-7, a r^{2}-9, a r^{3}-5 \rightarrow$ A.P. $a_{2}-a_{1}=a_{3}-a_{2}$

$(a r-7)-(a-2)=\left(a r^{2}-9\right)-(a r-7)$

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

$a(r-1)(r-1)=-3 …(i)$

$a_{2}-a_{1}=a_{4}-a_{3}$

$(a r-7)-(a-2)=\left(a r^{3}-5\right)-\left(a r^{2}-9\right)$

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

$=a(r-1)\left(r^{2}-1\right)=-9 …(ii)$

ii/i

$\Rightarrow r+1=3$

$\Rightarrow r=2$

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

$a=-3$

$x_{1}=-3, x_{2}=-6, x_{3}=-12, x_{4}=-24$

$\frac{1}{24}\left(x_{1} \cdot x_{2} \cdot x_{3} \cdot x_{4}\right)=216$