અહીં  $S_{2n} = 2S_n$

$\therefore \,\,\frac{2n}{2}\,(2a\,+\,(2n\,-\,1)\,d)\,\,=\,\,3\,\frac{n}{2}\,(2a\,+\,(n\,-\,1)\,d)\,\,\,\,$

$\therefore \,\,2a\,+\,(2n\,-\,1)\,d\,\,=\,\,\frac{3}{2}\,\,(2a\,+\,(n\,-\,1)\,d)$

$\therefore \,\,(2a\,+\,2nd\,-\,d)\,\,3\,\,(2a\,+\,nd\,-\,d)\,$

$\therefore \,\,\,\,2a\,\,=\,\,d\,\,(n\,+\,1)$

હવે,$\frac{{{S}_{3n}}}{{{S}_{n}}}\,\,=\,\,\frac{\frac{3n}{2}\,\left[ 2a\,+\,(3n\,-\,1)\,d \right]}{\frac{n}{2}\,\left[ 2a\,+\,(n\,-\,1)\,d \right]}$

$=\,\,\frac{3\,\left[ d\,(n\,+\,1)\,+\,(3n\,-\,1)\,d \right]}{d\,(n\,+\,1)\,+\,(n\,-\,1)\,d}$

$=\,\,\frac{3d\,(4n)}{d\,(2n)}\,\,=\,\,6$