NCERT Question

$\mathrm{For} a G.P, S_n=\frac{a\left(1-r^n\right)}{1-r}$$\mathrm{sum} \mathrm{of} (n+1) \mathrm{terms} \mathrm{to} \left(2n\right) \mathrm{terms} = S_{2n}-S_n$$S_{2n}-S_n=\frac{a\left(1-r^{2n}\right)}{1-r}-\frac{a\left(1-r^n\right)}{1-r}$$=\frac{a\left(r^{{}^n}-r^{2n}\right)}{1-r}=\frac{{\mathrm{ar}}^n\left(1-r^n\right)}{1-r}=r^n\frac{a\left(1-r^n\right)}{1-r}=r^n.S_n$$\mathrm{Hence}, \frac{S_n}{S_{2n}-S_n}=\frac{1}{r^n}$">