Medium MCQ +4 / -1 PYQ · JEE Mains 2025

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

  1. A $\frac{1}{7}$
  2. B 1
  3. C $\frac{6}{7}$
  4. D 6 Correct answer

Solution

<p>$$\begin{aligned} & \text { Let } S=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2 \alpha)+\ldots \\ & \frac{1}{7} S=\frac{1}{7}(5)+\frac{1}{7^2}(5+\alpha)+\ldots \infty \end{aligned}$$</p> <p>$$\begin{aligned} &\frac{6}{7}(S)=5+\frac{1}{7} \alpha\left(\frac{1}{1-\frac{1}{7}}\right)\\ &6=5+\frac{\alpha}{6} \Rightarrow \alpha=6 \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression

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