JEE Main 2024 — 9 April, Shift 2 — Question 28

$\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right)$

$-\left\{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right)\right\}=\frac{1}{2024}$

$\begin{aligned} \Rightarrow & \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) \end{aligned}$

$\begin{aligned} -\left\{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\ldots+\frac{1}{2023}\right. & \left.-\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2022}\right)\right\}=\frac{1}{2024}\end{aligned}$

$\begin{aligned} \Rightarrow & \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) \end{aligned}$

$\begin{aligned} -\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2023}\right) & +\frac{1}{2024}+\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{1011}\right)=\frac{1}{2024} \end{aligned}$

$\Rightarrow \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}$

$=\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023}$

$\Rightarrow \alpha=1011$