Question

How many terms of the sequence \[\sqrt{3}, 3, 3\sqrt{3},\]  ... must be taken to make the sum \[39 + 13\sqrt{3}\] ?

Answer 1

Here,a = \[\sqrt{3}\] Common ratio,r = \[\sqrt{3}\]

Sum of n terms, S_n = \[39 + 3\sqrt{3}\]

\[S_n = \sqrt{3}\left( \frac{\left( \sqrt{3} \right)^n - 1}{\sqrt{3} - 1} \right) \]

\[ \Rightarrow 39 + 13\sqrt{3} = \frac{\sqrt{3}}{\left( \sqrt{3} - 1 \right)}\left\{ \left( \sqrt{3} \right)^n - 1 \right\}\]

\[ \Rightarrow \left( \sqrt{3} \right)^n - 1 = \frac{\left( 39 + 13\sqrt{3} \right)\left( \sqrt{3} - 1 \right)}{\sqrt{3}}\]

\[ \Rightarrow \left( \sqrt{3} \right)^n = 1 + 26\]

\[ \Rightarrow \left( \sqrt{3} \right)^n = 27 \]

\[ \Rightarrow \left( \sqrt{3} \right)^n = \left( \sqrt{3} \right)^6 \]

\[ \therefore n = 6\]