Q33 / 75 · JEE Main 2025 (24 Jan Shift 2) · Mathematics

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

  • A
    $\frac{6}{7}$
  • Correct answer
    B
    6
  • C
    $\frac{1}{7}$
  • D
    1

Solution

$S=a+(a+d) r+(a+2 d) r^2+\ldots$ Then $S=\frac{a}{1-r}+\frac{d r}{(1-r)^2},|r| < 1$ Since, $r=\frac{1}{7}$ and $a=5, d=\alpha$ $\begin{aligned} & 7=\frac{5}{1-\frac{1}{7}}+\frac{\alpha \cdot \frac{1}{7}}{\left(1-\frac{1}{7}\right)^2} \\ & \Rightarrow \alpha=6 \end{aligned}$

Hence, the correct answer is option: 'B'.

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