Question

Which term of the G P 3,root3,3root3, ---------- is 729​

Answer 1

EXPLANATION.

=> Series are = 3,√3,3√3 ........ 729

$\sf : \implies \: general \: term \: of \: an \: gp \\ \\ \sf : \implies \: t_{n} = ar {}^{n - 1} \\ \\ \sf : \implies \: first \: term \: = 3 \\ \\ \sf : \implies \: common \: ratios \: = \frac{b}{a} = \frac{ \sqrt{3} }{3} = \frac{1}{ \sqrt{3} }$

$\sf : \implies \: 729 = 3 \times \dfrac{1}{ \sqrt{3} {}^{} } {}^{n - 1} \\ \\ \sf : \implies \: {3}^{6} = \sqrt{3} {}^{(n - 1)} \\ \\ \sf : \implies \: \sqrt{3} {}^{(12)} = \sqrt{3} {}^{(n - 1)} \\ \\ \sf : \implies \: 12 = n - 1 \\ \\ \sf : \implies \: n \: = 13$

Answer 2

Answer:

REFER TO THE ATTACHMENT

HOPE IT HELPS :)