2. In an infinite arithmetic progression with a positive difference, the numbers 25, 41, 65 are included (not necessarily consecutively). Is the number 2017 a member of this sequence? Justify your answer.
2. Since $25, 41, 65$ are terms of an arithmetic progression, then $25=a_{1}+k d ; 41=a_{1}+n d ; 65=a_{1}+m d$, where $k, n, m$ are natural numbers. From these three equations, it follows that $16=(n-k) d, 24=(m-n) d$. From these two equations, we get:

$8=(m-2 n+k) d$. Since $2017=1+2016$, and $2016=252 \cdot 8=3 \cdot 8+249 \cdot 8=24+$ $+249 \cdot 8=24+249(m-2 n+k) d$, then $2017=1+24+249(m-2 n+k) d=25+$ $+249(m-2 n+k) d=a_{1}+k d+249(m-2 n+k) d=a_{1}+(249 m-498 n+250 k) d$ or $2017=a_{1}+l d$, where $l-$ is a positive natural number. Therefore, 2017 is a term of this sequence.

Answer: Yes, it is.