Let $a_1, a_2, a_3, a_4, a_5, a_6$ be the first six terms of the G.P. According to question, sum of first three terms = 16 ⇒ $a_1 + a_2 + a_3$ = 16 ....(i) Sum of next three terms = 128 ⇒ $a_4 + a_5 + a_6$ = 128 ....(ii) Let $S_6$ be the sum of first six terms i.e., $S_6$ = 16 +128 = 144 [from (i) and (ii)] Let a be the first term and r be the common ratio, then $S_6$ = $\frac{a(1-r^6)}{1-r}$ ⇒ 144 = $\frac{a(1-r^6)}{1-r}$ ... (iii) and $S_3$ = $\frac{a(1-r^3)}{1-r}$ ⇒ 16 = $\frac{a(1-r^3)}{1-r}$ ... (iv) ∴ $\frac{S_6}{S_3}$ = $\frac{-r^6}{1-r^3}$ = $\frac{144}{16}$ [from (iii) and (iv)] ⇒ $\frac{(1-r^3)(1+r^3)}{1-r^3}$ = 9 ⇒ 1 + $r^3$ = 9 ⇒ $r^3$ = 8 ⇒ r = 2 ∴ $S_3$ = $\frac{a(1-r^3)}{1-r}$ ⇒ 16 = $\frac{a(1-8)}{1-2}$ = $\frac{7a}{1}$ ⇒ a = $\frac{16}{7}$ ∴ $S_n$ = $\frac{a(1-rn)}{1-r}$ = $\frac{16}{7}\frac{1-2^n}{1-2}$ = $\frac{16}{7}(2^n-1)$. Hence a = $\frac{16}{7}$ , r = 2 and $S_n$ = $\frac{16}{7}(2^n-1)$