If 7=5+1/7(5+α)+frac 1 7 2 (5+2 α)+frac 1 7 3 (5+3 α)+ \ \ \ \ \ ∞ , then the value of α is :

Mathematics · Sequence and Series

JEE Main 2025 — 24 January, Evening Shift — Question 8

If $7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^{2}}(5+2 \alpha)+\frac{1}{7^{3}}(5+3 \alpha)+$ $\_\_\_\_\_$ $\infty$ , then the value of $\alpha$ is :

Let $S=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^{2}}(5+2 \alpha)+\ldots$

$\frac{1}{7} S=\frac{1}{7}(5)+\frac{1}{7^{2}}(5+\alpha)+\ldots \infty$

$\frac{6}{7}(S)=5+\frac{1}{7} \alpha\left(\frac{1}{1-\frac{1}{7}}\right)$

$6=5+\frac{\alpha}{6}$

$\Rightarrow \alpha=6$ .

Answer key and solution verified before publishing.

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