JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Sequences and Series

AP${}_1$: $3,7,11,\dots ,403$ (common difference 4); AP${}_2$: $2,5,8,\dots ,404$ (common difference 3). Common terms form an AP with common difference $\text{lcm}(4,3)=12$, first common term $11$: $11,23,35,\dots$
Largest term of this AP not exceeding $403$ (the smaller upper bound): $11+12(n-1)\le 403\Rightarrow n-1\le 32.67\Rightarrow n-1=32$, giving last term $11+12(32)=395$ (checking: $395\equiv 3(\mathrm{mod}4)$ and $395\equiv 2(\mathrm{mod}3)$, so it does belong to both original APs), and $n=33$ terms in total.