Let a 1 , a 2 , a 3 , . . . . . . be a G.P. of increasing positive numbers. Let the sum of its 6 t h and 8 t h terms be 2 and the product of its 3 r d and 5 t h terms be 1 9 . Then 6 ( a 2 + a 4 ) ( a 4 + a 6 ) is equal to [2

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Solution

Q.
Let $a_{1}, a_{2}, a_{3}, . . . . . .$ be a G.P. of increasing positive numbers. Let the sum of its $6^{t h}$ and $8^{t h}$ terms be 2 and the product of its $3^{r d}$ and $5^{t h}$ terms be $\frac{1}{9} .$ Then $6 (a_{2} + a_{4}) (a_{4} + a_{6})$ is equal to [2023]

1 $2 \sqrt{2}$

2 2

3 3

4 $3 \sqrt{3}$

Ans.
(3)

$Given, a_{6} + a_{8} = 2$

$\Rightarrow a r^{5} (1 + r^{2}) = 2 ...(i)$

   $a_{3} \times a_{5} = \frac{1}{9} (Given)$

$\Rightarrow a r^{2} \times a r^{4} = \frac{1}{9} \Rightarrow {(a r^{3})}^{2} = \frac{1}{9} \Rightarrow a r^{3} = \pm \frac{1}{3} ...(ii)$

$Now, by (i) and (ii), we get$

$a r^{3} r^{2} (1 + r^{2}) = 2 \Rightarrow (\pm \frac{1}{3}) r^{2} (1 + r^{2}) = 2 \Rightarrow \frac{r^{2} (1 + r^{2})}{3} = 2$

$\Rightarrow r^{4} + r^{2} - 6 = 0 \Rightarrow r^{2} = 2 \Rightarrow r = \sqrt{2}$

   $a (2 \sqrt{2}) = \frac{1}{3}$ $(By (ii))$

   $a = \frac{1}{6 \sqrt{2}}$

$∴$ $6 (a_{2} + a_{4}) (a_{4} + a_{6}) = 6 (a r + a r^{3}) (a r^{3} + a r^{5})$

$= 6 a r (1 + r^{2}) a r^{3} (1 + r^{2}) = 6 a^{2} r^{4} {(1 + r^{2})}^{2}$

$= 6 (\frac{1}{72}) \cdot 4 {(1 + 2)}^{2} = 3$

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