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If < an > is an A.P. Such that a 4 a 7 = 2 3 , Find a 6 a 8 . - Mathematics

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प्रश्न

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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उत्तर

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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अध्याय 19: Arithmetic Progression - Exercise 19.2 [पृष्ठ १२]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 19 Arithmetic Progression
Exercise 19.2 | Q 23 | पृष्ठ १२

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