Let a1, a2, ..........., a21 be an AP such that $\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = {4 \over 9}}$. If the sum of this AP is 189, then a6a16 is equal to :
Solution
$$\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = \sum\limits_{n = 1}^{20} {{1 \over {{a_n}({a_n} + d)}}} } $$<br><br>$$ = {1 \over d}\sum\limits_{n = 1}^{20} {\left( {{1 \over {{a_n}}} - {1 \over {{a_n} + d}}} \right)} $$<br><br>$$ \Rightarrow {1 \over d}\left( {{1 \over {{a_1}}} - {1 \over {{a_{21}}}}} \right) = {4 \over 9}$$ (Given)<br><br>$$ \Rightarrow {1 \over d}\left( {{{{a_{21}} - {a_1}} \over {{a_1}{a_{21}}}}} \right) = {4 \over 9}$$<br><br>$$ \Rightarrow {1 \over d}\left( {{{{a_1} + 20d - {a_1}} \over {{a_1}{a_2}}}} \right) = {4 \over 9} \Rightarrow {a_1}{a_2} = 45$$ .... (1)<br><br>Now sum of first 21 terms = ${{21} \over 2}(2{a_1} + 20d) = 189$<br><br>$\Rightarrow$ a<sub>1</sub> + 10d = 9 ..... (2)<br><br>For equation (1) & (2) we get<br><br>a<sub>1</sub> = 3 & d = ${3 \over 5}$<br><br>or a<sub>1</sub> = 15 & d = $- {3 \over 5}$<br><br>So, a<sub>6</sub> . a<sub>16</sub> = (a<sub>1</sub> + 5d) (a<sub>1</sub> + 15d)<br><br>$\Rightarrow$ a<sub>6</sub>a<sub>16</sub> = 72<br><br>Option (b)
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Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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