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Question
Find :
nth term of the G.P.
\[\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, . . .\]
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Solution
Here,
\[\text { First term, } a = \sqrt{3}\]
\[\text { Common ratio, } r = \frac{a_2}{a_1} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{3}\]
\[ \therefore \text { nth term } = a_n = a r^{(n - 1)} = \sqrt{3} \left( \frac{1}{3} \right)^{n - 1} \]
\[\text { Thus, the nth term of the given GP is } \sqrt{3} \left( \frac{1}{3} \right)^{n - 1} .\]
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