JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Sequences and Series

The product of the first three terms is $a\cdot ar\cdot a{r}^2=(ar{)}^3=1$, so $ar=1$. For four terms the sum of the reciprocals equals $\frac{S}{t_1{t}_4}=\frac{S}{a^2{r}^3}$, so $\frac{65/12}{a^2{r}^3}=\frac{65}{18}$ and $a^2{r}^3=\frac{3}{2}$. Since $a=\frac{1}{r}$, this reads $r=\frac{3}{2}$, hence $a=\frac{2}{3}$ and the progression is $\frac{2}{3},1,\frac{3}{2},\frac{9}{4}$, whose sum is indeed $\frac{65}{12}$. The third term is $\alpha =a{r}^2=\frac{3}{2}$, so $2\alpha =3$.