Let be a G.P. of increasing positive terms, If and , then is equal to : [2025]

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Solution

Q.
Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

1 812

2 526

3 784

4 628

Ans.
(3)

Here, r > 0 and $a_{1}, a_{2}, a_{3}, . . . . . > 0$ as G.P. has increasing positive terms.

$a_{1} a_{5} = 28$ (Given)

$\Rightarrow a (a r^{4}) = 28 = a^{2} r^{4} = 28$ ... (i)

Also, $a_{2} + a_{4} = 29$

$\Rightarrow a r + a r^{3} = 29$

$\Rightarrow a r (1 + r^{2}) = 29$ ... (ii)

From (i) and (ii), we get

$\frac{r^{2}}{{(1 + r^{2})}^{2}} = \frac{28}{{(29)}^{2}}$

$\Rightarrow 841 r^{2} = 28 r^{4} + 56 r^{2} + 28$

$\Rightarrow 28 r^{4} - 785 r^{2} + 28 = 0$

$\Rightarrow (r^{2} - 28) (28 r^{2} - 1) = 0$

$\Rightarrow r^{2} = 28 or \frac{1}{28}$

$\Rightarrow r = \sqrt{28}$ $[∵ r \neq \frac{1}{28} as G . P . is increasing]$

$\Rightarrow a = \frac{1}{\sqrt{28}} and a_{6} = a r^{5} = \frac{1}{\sqrt{28}} \times {(\sqrt{28})}^{5} = 784$ .

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