Question
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 $A \cap B$, which are divisible by 3 , is :
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Solution
Sol. [Key Concept: Common terms of the two A.P.s form an A.P. with common difference $35$ (the lcm of $5$ and $7$); count those divisible by 3.]
$\ {AP}_{1}$ Set $A = \{ 1,6,11,16\ldots\ldots\ldots\ldots 101$ terms $\}$
${AP}_{2}$ Set $B = \{ 9,16$
71 terms
$$D = L.C.M\left\{ d_{1},{\text{ }d}_{2} \right\} = 35 $$$1^{\text{st}\text{~}}$ Common term is 16
$${16 + (n - 1)35 \leq 499 }{n \leq 14.8 }{\Rightarrow n = 14 }$$$A \cap B = \{ 16,51,86,121,156,191,226,261,296$, 331, 366,401,436,471}
terms divisible by ' 3 '
$$= \{ 51,156,261,366,471\} $$
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