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Question
Find:
the 10th term of the G.P.
\[- \frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9}, . . .\]
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Solution
Here,
\[\text { First term }, a = \frac{- 3}{4} \]
\[\text { Common ratio, } r = \frac{a_2}{a_1} = \frac{\frac{1}{2}}{- \frac{3}{4}} = - \frac{2}{3}\]
\[ \therefore 10th \text { term }= a_{10} = a r^{(10 - 1)} = \left( \frac{- 3}{4} \right) \left( \frac{- 2}{3} \right)^9 = \frac{1}{2} \left( \frac{2}{3} \right)^8 \]
\[\text { Thus, the 10th term of the given GP is } \frac{1}{2} \left( \frac{2}{3} \right)^8 .\]
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