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Question
Find the 4th term from the end of the G.P.
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Solution
\[\text { Here, first term, } a = \frac{2}{27}\]
\[\text { Common ratio}, r = \frac{a_2}{a_1} = \frac{\frac{2}{9}}{\frac{2}{27}} = 3 \]
\[\text { Last term, } l = 162\]
\[\text { After reversing the given G . P . , we get another G . P . whose first term is l and common ratio is } \frac{1}{r} . \]
\[ \therefore 4th \text { term from the end } = l \left( \frac{1}{r} \right)^{4 - 1} = (162) \left( \frac{1}{3} \right)^3 = 6\]
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