If , then the value of is : [2025]

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Solution

Q.
If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7^{2}} (5 + 2 α) + \frac{1}{7^{3}} (5 + 3 α) + . . . . . \infty$ , then the value of $α$ is : [2025]

1 $\frac{6}{7}$

2 $\frac{1}{7}$

3 6

4 1

Ans.
(3)

Given: $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7^{2}} (5 + 2 α) + \frac{1}{7^{3}} (5 + 3 α) + . . . . . \infty$

Let $k = 5 + \frac{1}{7} (5 + α) + \frac{1}{7^{2}} (5 + 2 α) + \frac{1}{7^{3}} (5 + 3 α) + . . . . . \infty$

$\Rightarrow \frac{k}{7} = \frac{5}{7} + \frac{1}{7^{2}} (5 + α) + \frac{1}{7^{3}} (5 + 2 α) + . . . . . \infty$

$\Rightarrow k - \frac{k}{7} = 5 + \frac{1}{7} α (\frac{1}{1 - \frac{1}{7}})$

$\Rightarrow \frac{6 k}{7} = 5 + \frac{α}{7} \times \frac{7}{6}$

Now, $k - \frac{k}{7} = 5 + \frac{α}{7} + \frac{α}{7^{2}} + \frac{α}{7^{3}} + . . . \infty$

$\Rightarrow \frac{6 k}{7} = 5 + \frac{α}{6}$

$\Rightarrow \frac{6 \times 7}{7} = 5 + \frac{α}{6}$ [ $∵$ k = 7]

$\Rightarrow \frac{α}{6} = 1 \Rightarrow α = 6$ .

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