If a1, a2, a3 ...... and b1, b2, b3 ....... are A.P., and a1 = 2, a10 = 3, a1b1 = 1 = a10b10, then a4 b4 is equal to -
Solution
<p>a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> .... are in A.P. (Let common difference is d<sub>1</sub>)</p>
<p>b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub> .... are in A.P. (Let common difference is d<sub>2</sub>)</p>
<p>and a<sub>1</sub> = 2, a<sub>10</sub> = 3, a<sub>1</sub>b<sub>1</sub> = 1 = a<sub>10</sub>b<sub>10</sub></p>
<p>$\because$ a<sub>1</sub>b<sub>1</sub> = 1</p>
<p>$\therefore$ b<sub>1</sub> = ${1 \over 2}$</p>
<p>a<sub>10</sub>b<sub>10</sub> = 1</p>
<p>$\therefore$ b<sub>10</sub> = ${1 \over 3}$</p>
<p>Now, a<sub>10</sub> = a<sub>1</sub> + 9d<sub>1</sub> $\Rightarrow$ d<sub>1</sub> = ${1 \over 9}$</p>
<p>b<sub>10</sub> = b<sub>1</sub> + 9d<sub>2</sub> $\Rightarrow$ d<sub>2</sub> = ${1 \over 9}$$\left[ {{1 \over 3} - {1 \over 2}} \right]$ = $-$ ${{1 \over {54}}}$</p>
<p>Now, a<sub>4</sub> = 2 + ${{3 \over {9}}}$ = ${{7 \over {3}}}$</p>
<p>b<sub>4</sub> = ${{1 \over {2}}}$ $-%$ ${{3 \over {54}}}$ = ${{4 \over {9}}}$</p>
<p>$\therefore$ a<sub>4</sub>b<sub>4</sub> = ${{28 \over {27}}}$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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