Question 217, Determine the smallest possible value of the largest term in an arithmetic sequence composed of seven distinct prime numbers.

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Question 217, Solution: Let the seven prime numbers be $p_{1}=p, p_{2}=p+d, p_{3}=p+2 d, p_{4}=p+3 d, p_{5}=p+4 d, p_{6}=p+5 d, p_{7}=p+6 d$.
If $d$ is odd, then $p_{6}$ and $p_{7}$ must be one odd and one even, and both greater than 2, which cannot both be prime. Therefore, $d$ must be even.
If $3 \nmid d$, then $p_{5}, p_{6}, p_{7}$ form a complete residue system modulo 3, one of which must be divisible by 3, and all are greater than 3, which cannot all be prime. Therefore, $3 \mid d$.
If $5 \nmid d$, then $p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ form a complete residue system modulo 5, one of which must be divisible by 5, and all are greater than 5, which cannot all be prime. Therefore, $5 \mid d$.
Combining the above three aspects, we know that $30 \mid d$. Let $d=30k$.
(1) If $7 \nmid d$, then $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ form a complete residue system modulo 7, one of which must be divisible by 7, so this number can only be 7, and it can only be $p_{1}=7$. Thus, $p_{1}=7+30k(i-1)$.
When $k=1$, $p_{7}=187=11 \times 17$, not a prime, does not meet the condition;
When $k=2$, $p_{4}=187=11 \times 17$, not a prime, does not meet the condition;
When $k=3$, $p_{3}=187=11 \times 17$, not a prime, does not meet the condition;
When $k=4$, $p_{3}=247=13 \times 19$, not a prime, does not meet the condition;
When $k=5$, $p_{1}=7, p_{2}=157, p_{3}=307, p_{4}=457, p_{5}=607, p_{6}=757, p_{7}=907$, all are prime.
Therefore, when $7 \nmid d$, the smallest value of $p_{7}$ that meets the condition is 907.
(2) If $7 \mid d$, then $210 \mid d$, thus $p_{7}>6d \geq 6 \times 210=1260>907$.
In summary, the smallest possible value of the largest term in the arithmetic sequence that meets the condition is $907$.