[shahkb4]

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 is $\mathrm{A} \cap \mathrm{B}$, which are divisible by 3 , is
[JEE MAIN 02 April 2026 S I]
(1) 4
(2) 5
(3) 6
(4) 7

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[shahkb4]

Solution

\[ \mathrm{AP}_{1} \text { Set } \mathrm{A}=\{1,6,11,16 \ldots \ldots \ldots \ldots\}, \mathrm{n}(\mathrm{~A})=101 \]

\[ \begin{aligned} & \mathrm{T}_{101}^{\mathrm{A}}=1+100 \times 5=501 \\ & \mathrm{AP}_{2} \text { Set } \mathrm{B}=\{9,16 \ldots \ldots \ldots\} \mathrm{N}(\mathrm{~B})=71 \\ & \mathrm{~T}_{71}^{\mathrm{B}}=9+70 \times 7=499 \end{aligned} \]

Common A.P., common difference $\mathrm{D}=\mathrm{L} . \mathrm{C} . \mathrm{M}\left\{\mathrm{d}_{1}, \mathrm{~d}_{2}\right\}=35$;
$1^{\text {st }}$ Common term is 16

\[ \begin{aligned} & 16+(n-1) 35 \leq 499 \Rightarrow n \leq 14.8 \Rightarrow n=14 \\ & T_{n}^{A \cap B}=16+(n-1) 35=35 n-19 \text { is divisible by } 3 \\ & =3(11 n-6)+2 n-1 \text { is divisible by } 3 \\ & 2 n-1 \text { must be divisible by } 3 \\ & n \in\{2,5,8,11,14\} \therefore 5 \text { such terms } \end{aligned} \]

[CHAPTER M08 PERMUTATION \& COMBINATION TOPIC T3 ]

[shahkb4]

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