JEE Main 2025 — 22 January, Evening Shift — Question 5

$a_{1}, a_{2}, a_{3}, \ldots \ldots, a_{2 k}$

$\sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}-1}=40, \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}}=55, \mathrm{a}_{2 \mathrm{k}}-\mathrm{a}_{1}=27$

$\frac{\mathrm{k}}{2}\left[2 \mathrm{a}_{1}+(\mathrm{k}-1) 2 \mathrm{~d}\right]=40, \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_{2}+(\mathrm{k}-1) 2 \mathrm{~d}\right]=55$,

$\mathrm{d}=\frac{27}{2 \mathrm{k}-1}$

$\mathrm{a}_{1}=\frac{40}{\mathrm{k}}-(\mathrm{k}-1) \mathrm{d}=\frac{55}{\mathrm{k}}-\mathrm{kd}$

$\mathrm{d}=\frac{15}{\mathrm{k}} \Rightarrow \frac{27}{2 \mathrm{k}-1}=\frac{15}{\mathrm{k}} \Rightarrow 9 \mathrm{k}=10 \mathrm{k}-5$

$\therefore \mathrm{k}=5$.