JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Sequences and Series

Take the four terms as $a,ar,a{r}^2,a{r}^3$ with $a,r>0$. The product gives $a^4{r}^6=1296$, so $a^2{r}^3=36$ and $a=\frac{6}{r^{3/2}}$. The sum condition $a(1+r+r^2+r^3)=126$ then becomes $r^{-3/2}+r^{-1/2}+r^{1/2}+r^{3/2}=21$. Setting $A=r^{1/2}+r^{-1/2}$ and using $A^3=r^{3/2}+r^{-3/2}+3A$, the equation reduces to $A^3-2A-21=0$, whose only real root is $A=3$. Then $r^{1/2}+r^{-1/2}=3$ gives $r^2-7r+1=0$. The two admissible common ratios are the roots of this quadratic, and their sum is $7$.