Let be a G.P. of increasing positive numbers. 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 numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to [2025]

1 131

2 129

3 128

4 130

Ans.
(2)

Let $a$ be the first term and $r$ be the common ratio of GP respectively.

Given, $a_{3} a_{5} = 729 \Rightarrow a r^{2} \cdot a r^{4} = 729$

$\Rightarrow a^{2} r^{6} = 729 \Rightarrow a r^{3} = 27$ ... (i)

and $a_{2} + a_{4} = a r + a r^{3} = \frac{111}{4}$

$\Rightarrow a r + 27 = \frac{111}{4} \Rightarrow a r = \frac{3}{4}$ ... (ii)

Now, divide equation (i) by equation (ii), we get

$\frac{a r^{3}}{a r} = \frac{27}{3 / 4}$

$\Rightarrow r^{2} = 36 \Rightarrow r = 6$ [ $∵$ G.P. is an increasing series]

Substitute $r$ = 6 in equation (ii), we get

$a = \frac{1}{8}$

Now, $24 (a_{1} + a_{2} + a_{3})$

$= 24 (a + a r + a r^{2})$

$= 24 a (1 + r + r^{2})$

$= 24 \times \frac{1}{8} (1 + 6 + 36)$

= 3(43) = 129.

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