Question

which term of G.P. :√3,3√3,......is 729?

Answer 1

hello Meenakshi

Here is your answer

◆ Given that

a = √3

And common ratio r = 3 /√3
= √3

As we know that nth term of GP
$a_{n} = a {r}^{n - 1} \\ \\ nth \: term \: is \: given \: as \: 729 \\ \\ 729 = \sqrt{3} \times { \sqrt{3} }^{n - 1} \\ 729 \div \sqrt{3} = { \sqrt{3} }^{n - 1} \\ { \sqrt{3} }^{12} \div \sqrt{3} = { \sqrt{3} }^{n - 1} \\ { \sqrt{3} }^{11} = { \sqrt{3} }^{n - 1} \\ \\ by \: comparison \\ 11 = n - 1 \\ n = 12$

Hence, the 12 th value will be 729 of the GP


hope it helps
jerri