Suppose such an arithmetic progression exists, with common difference $d$. We assume wlog it is increasing. We have $2|d$, since otherwise every other term would be even, and the only even prime is $2$. We also have $3|d$, since otherwise every third term would be divisible by 3, and the only such prime is $3$. We also have $5|d$, since otherwise at least two terms would be divisible by 5, but the only such prime is $5$. If 7 does not divide $d$, then 7 is a member of the arithmetic progression, as at least one of the 11 members must be divisible by 7. But 7 being the fourth prime, it is at most the fourth member of the progression. Hence the eleventh member is $7+7d$, which is divisible by 7 and not 7. Hence $7|d$. If $11|d$ then $2310|d$, and then clearly the last term is greater than 20000. Therefore 11 does not divide d, and exactly one member of the progression is divisible by 11. This must be 11. Clearly 11 is the first term. Now $d<2000$ and $210|d$. Also, $11+d$ is prime. By calculation, this narrows $d$ down to 420,630,1050,1470,1890. After a lot more calculation, the fact that $11+2d$ is prime means that $d=1050$. But then $11+3d=3161$ is not prime, and so no such $d$ can exist. Contradiction.

More generally (cited e.g. by W. Sierpinski in [250 Problems of Elementary Number Theory]) Theorem (Thébault). If $n$ primes form an arithmetic progression, then its ratio is divisible by any of the primes $p< n$ (and if $n$ is itself a prime, then either it is the first term, or else the ratio is also divisible by $n$).

Oopsies. Wrong place.

Also, been done here: E 40 - PEN