Question

find the 9th term and the n th term of the series 1 + √3 + 3 + 3√ 3 +.....​

Answer 1

     

This is a GP of first term 1 and common ratio √3.

a = 1

r = √3

nth term =

$\Rightarrow\ ar^{n-1} \\ \\ \Rightarrow\ 1 \times (\sqrt{3})^{n-1} \\ \\ \Rightarrow\ (\sqrt{3})^{n-1} \\ \\ \Rightarrow\ (3^{\frac{1}{2}})^{n-1} \\ \\ \Rightarrow\ 3^{\frac{n-1}{2}}$

9th term =

$\Rightarrow\ 3^{\frac{9-1}{2}} \\ \\ \Rightarrow\ 3^{\frac{8}{2}} \\ \\ \Rightarrow\ 3^4 \\ \\ \Rightarrow\ \bold{81}$

Hope this helps.

Thank you. :-))

       

Answer 2

Common ratio of GP, r = √3

T(n) = $ar^{n-1}$

T(9) = $1(\sqrt{3}) ^ {9-1}$

= $(\sqrt{3}) ^ {8}$

= $3^{\frac{8}{2} }$

= $3^{4}$ = 81.

nth term, T(n) = $1(\sqrt{3}) ^ {n-1}$

= $(\sqrt{3})^{n-1}$