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Question
How many terms of the sequence \[\sqrt{3}, 3, 3\sqrt{3},\] ... must be taken to make the sum \[39 + 13\sqrt{3}\] ?
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Solution
Here,a = \[\sqrt{3}\] Common ratio,r = \[\sqrt{3}\]
Sum of n terms, Sn = \[39 + 3\sqrt{3}\]
\[S_n = \sqrt{3}\left( \frac{\left( \sqrt{3} \right)^n - 1}{\sqrt{3} - 1} \right) \]
\[ \Rightarrow 39 + 13\sqrt{3} = \frac{\sqrt{3}}{\left( \sqrt{3} - 1 \right)}\left\{ \left( \sqrt{3} \right)^n - 1 \right\}\]
\[ \Rightarrow \left( \sqrt{3} \right)^n - 1 = \frac{\left( 39 + 13\sqrt{3} \right)\left( \sqrt{3} - 1 \right)}{\sqrt{3}}\]
\[ \Rightarrow \left( \sqrt{3} \right)^n = 1 + 26\]
\[ \Rightarrow \left( \sqrt{3} \right)^n = 27 \]
\[ \Rightarrow \left( \sqrt{3} \right)^n = \left( \sqrt{3} \right)^6 \]
\[ \therefore n = 6\]
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