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 :
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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