Let A1, A2, A3, ....... be an increasing geometric progression of positive real numbers. If A1A3A5A7 = ${1 \over {1296}}$ and A2 + A4 = ${7 \over {36}}$, then the value of A6 + A8 + A10 is equal to
Solution
<p>$${{{A_4}} \over {{r^3}}}.\,{{{A_4}} \over r}.\,{A_4}r\,.\,{A_4}{r^3} = {1 \over {1296}}$$</p>
<p>${A_4} = {1 \over 6}$</p>
<p>${A_2} = {7 \over {36}} - {1 \over 6} = {1 \over {36}}$</p>
<p>So ${A_6} + {A_8} + {A_{10}} = 1 + 6 + 36 = 43$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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