JEE Main 2026 — Apr 2, Shift 1 — Mathematics solution

JEE Main2026Apr 2, Shift 1Mathematics

Let $A$ be the set of first $101$ terms of an A.P., whose first term is $1$ and the common difference is $5$ and let $B$ be the set of first $71$ terms of an A.P., whose first term is $9$ and the common difference is $7$ . Then the number of elements in $A \cap B$ , which are divisible by $3$ , is :

A
$4$
B
$5$
C
$6$
D
$7$

Solution

$A = {1, 6, 11, \dots}$ with last term $1 + 100 \cdot 5 = 501$ . $B = {9, 16, 23, \dots}$ with last term $9 + 70 \cdot 7 = 499$ .

Common terms form an A.P. with common difference $lcm (5, 7) = 35$ . The smallest common term is $16$ (since $16 \in A$ as $1 + 3 \cdot 5$ and $16 \in B$ as $9 + 7$ ).

Common terms: $16 + 35 (n - 1) \leq 499 \Rightarrow n \leq 14.8$ , so $14$ common terms:

$16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471$ .

Among these, the ones divisible by $3$ are $51, 156, 261, 366, 471$ — a total of $5$ terms.

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