Let AP(a; d) denote the set of all the terms | vedclass

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(D) The general terms of the three arithmetic progressions are $x = 3m + 1$,$x = 5n + 2$,and $x = 7k + 3$ for non-negative integers $m, n, k$.We need

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$157$

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(D) The general terms of the three arithmetic progressions are $x = 3m + 1$,$x = 5n + 2$,and $x = 7k + 3$ for non-negative integers $m, n, k$.We need to solve the system of congruences:$x \equiv 1 \pmod{3}$$x \equiv 2 \pmod{5}$$x \equiv 3 \pmod{7}$From $x \equiv 1 \pmod{3}$,$x = 3m + 1$.Substituting into the second congruence: $3m + 1 \equiv 2 \pmod{5}$ $\Rightarrow 3m \equiv 1 \equiv 6 \pmod{5}$ $\Rightarrow m \equiv 2 \pmod{5}$. So $m = 5n + 2$.Then $x = 3(5n + 2) + 1 = 15n + 7$.Substituting into the third congruence: $15n + 7 \equiv 3 \pmod{7}$ $\Rightarrow n + 0 \equiv 3 \pmod{7}$ $\Rightarrow n = 7k + 3$.Then $x = 15(7k + 3) + 7 = 105k + 45 + 7 = 105k + 52$.Thus,the first term $a = 52$ and the common difference $d = 	ext{lcm}(3, 5, 7) = 105$.Therefore,$a + d = 52 + 105 = 157$.

Let $AP(a; d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d > 0$. If $AP(1; 3) \cap AP(2; 5) \cap AP(3; 7) = AP(a; d)$,then $a + d$ equals:

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