JEE Main 2026 — Sequences and Series Question with Solution

The set $A$ consists of the first 101 terms of an A.P. with first term $a_1=1$ and common difference $d_1=5$.

The terms of $A$ are $1,6,11,16,\dots ,1+(101-1)\times 5=501$.

The set $B$ consists of the first 71 terms of an A.P. with first term $b_1=9$ and common difference $d_2=7$.

The terms of $B$ are $9,16,23,30,\dots ,9+(71-1)\times 7=499$.

The first common term of $A$ and $B$ is $16$.

The common difference of the terms in $A\cap B$ is the least common multiple of $d_1$ and $d_2$, which is $\text{lcm}(5,7)=35$.

The general term of $A\cap B$ is given by $T_k=16+35k$, where $k\ge 0$ is an integer.

Since the maximum possible value in $A\cap B$ cannot exceed the maximum value in $B$ (which is $499$), we have:

$16+35k\le 499$

$35k\le 483$

$k\le 13.8$

Thus, the possible values for $k$ are $0,1,2,\dots ,13$.

We require the terms in $A\cap B$ to be divisible by $3$. Therefore:

$16+35k\equiv 0(\mathrm{mod}3)$

$1+2k\equiv 0(\mathrm{mod}3)$

$2k\equiv 2(\mathrm{mod}3)$

$k\equiv 1(\mathrm{mod}3)$

The values of $k$ in the range $0\le k\le 13$ that satisfy $k\equiv 1(\mathrm{mod}3)$ are $1,4,7,10,13$.

There are exactly $5$ such values of $k$, which correspond to $5$ elements in $A\cap B$ that are divisible by $3$.

Answer: $5$