JEE Main 2026 — 24 January, Morning Shift — Question 15

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}_{1}+\mathrm{a}_{\mathrm{n}}\right]=\frac{525}{2}, \mathrm{~d}=\frac{-3}{4}$

$$\begin{aligned} & \frac{n}{2}\left[a_{1}+\frac{a_{1}}{4}\right]=\frac{525}{2} \end{aligned}$$

$\frac{5 a_{1} n}{4}=525$

$a_{1} n=420$

$a_{n}=a_{1}+(n-1)\left(\frac{-3}{4}\right)$

$\Rightarrow \frac{-3}{4} a_{1}=\left(\frac{-3}{4}\right)(n-1)$

$\Rightarrow a_{1}=n-1$

$n(n-1)=420$

$n^{2}-n-420=0$

$(n-21)(n+20)=0$

$n=21, a_{1}=20$

$\sum_{i=1}^{17} a_{i}=\frac{17}{2}\left[2 a_{1}+16 d\right]$

$=\frac{17}{2}\left[40+16\left(\frac{-3}{4}\right)\right]$

$=\frac{17}{2}[40-12]$

$=17 \times 14=238$.