Intersection of Arithmetic Progressions and Term Sum

To find the common terms of the three arithmetic progressions, let $x$ be a term present in all three sets $AP(1; 3)$, $AP(2; 5)$, and $AP(3; 7)$.

The elements of each AP can be represented by the following modular congruences: $x \equiv 1 \pmod 3$ $x \equiv 2 \pmod 5$ $x \equiv 3 \pmod 7$

Since the common differences $3$, $5$, and $7$ are pairwise coprime, the common difference $d$ of the intersection set $AP(a; d)$ is given by the Least Common Multiple (LCM) of the common differences: $d = \text{lcm}(3, 5, 7) = 3 \times 5 \times 7 = 105$

Now, we find the smallest positive integer $a$ (the first term of the intersection) that satisfies all three congruences simultaneously.

1. Combining the second and third congruences: $x \equiv 2 \pmod 5 \implies x = 5k + 2$ Substituting this into the third congruence: $5k + 2 \equiv 3 \pmod 7 \implies 5k \equiv 1 \pmod 7$ Multiplying both sides by $3$ (since $3 \times 5 = 15 \equiv 1 \pmod 7$): $15k \equiv 3 \pmod 7 \implies k \equiv 3 \pmod 7$ Thus, $k = 7m + 3$, which gives: $x = 5(7m + 3) + 2 = 35m + 17$ So, $x \equiv 17 \pmod{35}$.

2. Now, combining $x \equiv 17 \pmod{35}$ with the first congruence $x \equiv 1 \pmod 3$: $35m + 17 \equiv 1 \pmod 3$ Since $35 \equiv 2 \pmod 3$ and $17 \equiv 2 \pmod 3$: $2m + 2 \equiv 1 \pmod 3 \implies 2m \equiv -1 \equiv 2 \pmod 3 \implies m \equiv 1 \pmod 3$ Thus, $m = 3n + 1$, which gives: $x = 35(3n + 1) + 17 = 105n + 52$

Therefore, the smallest positive integer term common to all three progressions is $a = 52$.

Thus, the set of common terms is $AP(52; 105)$, which means: $a = 52 \quad \text{and} \quad d = 105$

Finally, we compute the required value: $a + d = 52 + 105 = 157$