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 \(A\) 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:
[JEE Main 2026, 2 Apr (Shift 1)]
\(5\)
First A.P.
Set \(\mathrm{A}=\{1,6,11,16 \ldots .101\) terms \(\}\)
Second A.P.
Set \(\mathrm{B}=\{9,16 \ldots . .71\) terms \(\}\)
\(\mathrm{D}=\mathrm{L} . \mathrm{C} . \mathrm{M}\left\{\mathrm{d}_1, \mathrm{~d}_2\right\}=35\)
\(1^{\text {st }}\) Common term is \(16\)
\(16+(\mathrm{n}-1) 35 \leq 499\)
\(\mathrm{n} \leq 14.8\)
\(\Rightarrow \mathrm{n}=14\)
\(A \cap B=\{16,51,86,121,156,191,226,261,296,331,366,401,436,471\}\)
Terms divisible by \(3=\{51,156,261,366,471\}\)
\(=5\) terms
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