Let A be the set of first 101 terms of an A.P | vedclass

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(B) The set $A$ consists of terms $a_n = 1 + (n-1)5 = 5n - 4$ for $n = 1, 2, \dots, 101$. The maximum value is $5(101) - 4 = 501$.The set $B$ consists

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$5$

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(B) The set $A$ consists of terms $a_n = 1 + (n-1)5 = 5n - 4$ for $n = 1, 2, \dots, 101$. The maximum value is $5(101) - 4 = 501$.The set $B$ consists of terms $b_m = 9 + (m-1)7 = 7m + 2$ for $m = 1, 2, \dots, 71$. The maximum value is $7(71) + 2 = 499$.For an element $x$ to be in $A \cap B$, we have $x = 5n - 4 = 7m + 2$, which implies $5n = 7m + 6$.Testing values for $m$: If $m=1, x=9$ (not in $A$). If $m=2, x=16$ ($16 = 5(4)-4$, so $16 \in A$).The common terms form an $A$.$P$. with common difference $	ext{lcm}(5, 7) = 35$. Thus, $x = 35k + 16$.We require $16 \le 35k + 16 \le 499$, which gives $0 \le k \le 13.8$, so $k \in \{0, 1, 2, \dots, 13\}$.We want $x$ to be divisible by $3$: $35k + 16 \equiv 2k + 1 \equiv 0 \pmod 3$.This implies $2k \equiv -1 \equiv 2 \pmod 3$, so $k \equiv 1 \pmod 3$.The possible values for $k$ are $1, 4, 7, 10, 13$.There are $5$ such values.

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