(a) Let 128 be the nth term of the given G.P. Here a = 2 and r = $\sqrt{2}$ Therefore, $ar^{n-1}$ = 128 ⇒ $2(\sqrt{2})^{n-1}$ = 128 ⇒ $(\sqrt{2})^{n-1}$ = 64 ⇒ $(2)^{\frac{n-1}{2}}$ = $(2)^{6}$ ∴ $\frac{n-1}{2}$ = 6 ⇒ n – 1 = 12 ⇒ n = 13. Hence, 128 is the 13th term of the given G.P. (b) Let 729 be the nth term of the given G.P. Here a = $\sqrt{3}$ and r = $\sqrt{3}$ Therefore, $ar^{n-1}$ = $729$ ⇒ $(\sqrt{3})(\sqrt{3})^{n-1}$ = 729 ⇒ $(\sqrt{3})(3)^{\frac{n-1}{2}}$ = 729 ⇒ $3^{\frac{n-1}{2}+\frac{1}{2}}$ = 729 ⇒ $3^{\frac{n}{2}}$ = $(3)^{6}$ ∴ $\frac{n}{2}$ = 6 ⇒ n = 12 Hence 729 is the 12th term of the given G.P. (c) Let $\frac{1}{19683}$ be the nth term of the given G.P. Here a = $\frac{1}{3}$ and r = $\frac{1}{3}$ ∴ $ar^{n-1}$ = $\frac{1}{19683}$ ⇒ $\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{n-1}$ = $\frac{1}{19683}$ ⇒ $\left(\frac{1}{3}\right)^{n}$ = $\left(\frac{1}{3}\right)^{9}$ ∴ n = 9 Hence, $\frac{1}{19683}$ is the 9th term of the given G.P.