Let { a k } and { b k } k , be two G.P.s with common ratios r 1 and r 2 respectively such that a 1 = b 1 = 4 and r 1 r 2 . Let c k = a k + b k , k . If c 2 = 5 and c 3 = 13 4 then k = 1 ? c k - ( 12 a 6 + 8 b 4 ) is equal to _______

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Solution

Q.
Let ${a_{k}}$ and ${b_{k}}$ $k \in ℕ,$ be two G.P.s with common ratios $r_{1}$ and $r_{2}$ respectively such that $a_{1} = b_{1} = 4$ and $r_{1} < r_{2} .$ Let $c_{k} = a_{k} + b_{k},$ $k \in ℕ$ . If $c_{2} = 5$ and $c_{3} = \frac{13}{4}$ then $\sum_{k = 1}^{\infty} c_{k} - (12 a_{6} + 8 b_{4})$ is equal to ________ . [2023]

Ans.
(9)

Given, $c_{k} = a_{k} + b_{k}$

$a_{1} = b_{1} = 4$

Now, $a_{2} = 4 r_{1}$    $a_{3} = 4 r_{1}^{2}$

   $b_{2} = 4 r_{2}$    $b_{3} = 4 r_{2}^{2}$

Now, $c_{2} = a_{2} + b_{2} = 5$

   $c_{3} = a_{3} + b_{3} = \frac{13}{4}$

$\Rightarrow r_{1} + r_{2} = \frac{5}{4} and r_{1}^{2} + r_{2}^{2} = \frac{13}{6}$

So, $r_{1} r_{2} = \frac{3}{8} which gives r_{1} = \frac{1}{2}, r_{2} = \frac{3}{4}$

$\sum_{k = 1}^{\infty} c_{k} - (12 a_{6} + 8 b_{4}) = \frac{4}{1 - r_{1}} + \frac{4}{1 - r_{2}} - (\frac{48}{32} + \frac{27}{2})$

$= 24 - 15 = 9$

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