JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Sequences and Series

Let the first term be $a$ and the ratio $r$ with $|r|<1$. The sum is $\frac{a}{1-r}=3$, and the sum of cubes is $\frac{a^3}{1-r^3}=\frac{27}{19}$. Cubing the first gives $\frac{a^3}{(1-r{)}^3}=27$, and dividing by the cube-sum relation yields $\frac{1-r^3}{(1-r{)}^3}=19$, i.e. $\frac{1+r+r^2}{(1-r{)}^2}=19$. This becomes $6r^2-13r+6=0$, factoring as $(2r-3)(3r-2)=0$, so $r=\frac{3}{2}$ or $r=\frac{2}{3}$; since the terms are positive with $|r|<1$, the ratio is $\frac{2}{3}$.