JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Sequences and Series

The set $A=\{1,6,11,\dots \}$ has terms $\equiv 1(\mathrm{mod}5)$ with last term $1+100\cdot 5=501$, and $B=\{9,16,23,\dots \}$ has terms $\equiv 2(\mathrm{mod}7)$ with last term $9+70\cdot 7=499$. A common term satisfies both congruences; the smallest is $16$, and such terms recur with period $\mathrm{lcm}(5,7)=35$, so they are $16+35t$ with $16+35t\le 499$, giving $t=0,1,\dots ,13$ (14 terms). For divisibility by $3$, $16+35t\equiv 1+2t\equiv 0(\mathrm{mod}3)$, i.e. $t\equiv 1(\mathrm{mod}3)$, so $t=1,4,7,10,13$. These give $51,156,261,366,471$, which is $5$ terms.