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Arithmetic Progression

If the number...

Question

If the numbers $3^{2 a - 1}, 14, 3^{4 - 2 a}$ $(0 < a < 1)$ are in A.P., then the value of $(2 a + 1) (3 a - 1)$ is

\frac{11}{2}

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\frac{7}{2}

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4

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1

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Solution

The correct option is D $1$
Given that $3^{2 a - 1}, 14, 3^{4 - 2 a}$ are in A.P.
$∴ 3^{2 a - 1} + 3^{4 - 2 a} = 28$
$\Rightarrow \frac{t}{3} + \frac{81}{t} = 28$ , where $t = 3^{2 a}$
$\Rightarrow t^{2} - 84 t + 243 = 0 \Rightarrow (t - 81) (t - 3) = 0$
$\Rightarrow t = 81$ or $t = 3$
$\Rightarrow 3^{2 a} = 3^{4}$ or $3^{2 a} = 3^{1}$
$\Rightarrow a = 2 or a = \frac{1}{2}$
But $0 < a < 1. ∴ a = \frac{1}{2}$
Hence, $(2 a + 1) (3 a - 1) = 2 \times \frac{1}{2} = 1$

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