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Question
If < an > is an A.P. such that \[\frac{a_4}{a_7} = \frac{2}{3}, \text { find }\frac{a_6}{a_8}\].
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Solution
Given:
< an > is an A.P.
\[\frac{a_4}{a_7} = \frac{2}{3}\]
\[ \Rightarrow \frac{a + \left( 4 - 1 \right)d}{a + \left( 7 - 1 \right)d} = \frac{2}{3} \]
\[ \Rightarrow \frac{a + 3d}{a + 6d} = \frac{2}{3}\]
\[ \Rightarrow 3(a + 3d) = 2(a + 6d) \]
\[ \Rightarrow 3a + 9d = 2a + 12d\]
\[ \Rightarrow a = 3d . . . . (i)\]
\[\therefore \frac{a_6}{a_8} = \frac{a + \left( 6 - 1 \right)d}{a + \left( 8 - 1 \right)d}\]
\[ \Rightarrow \frac{a_6}{a_8} = \frac{a + 5d}{a + 7d}\]
\[ \Rightarrow \frac{a_6}{a_8} = \frac{3d + 5d}{3d + 7d} \left( \text { From }(i) \right)\]
\[ \Rightarrow \frac{a_6}{a_8} = \frac{8d}{10d}\]
\[ \Rightarrow \frac{a_6}{a_8} = \frac{4d}{5d} = \frac{4}{5}\]
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