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Suppose three primes  
\[p_1, \; p_2, \; p_3\]
  with each prime greater than 3, form an arithmetic progression. The common difference must be a multiple of 2, since each prime is odd. If the common difference  
\[d\]
  is not a multiple of 3, so  
\[d=3k+1\]
  or  
\[3k+2\]
.  
\[p_1\]
  is not a multiple of 3 either, so  
\[p_1=3j+1\]
  or  
\[3j+2\]
.
\[d, \; p_1 \ p_2, \; p_3\]
\[p_2=p_1+d\]
\[p_3=p_1+2d\]
\[d=3k+1, \; p_1=3j+1\]
\[3k+3j+1+2\]
\[6k+3j+3\]
\[d=3k+2, \; p_1=3j+1\]
\[3k+3j+3\]
\[6k+3j+5\]
\[d=3k+1, \; p_1=3j+2\]
\[3k+3j+3\]
\[6k+3j+4\]
\[d=3k+2, \; p_1=3j+2\]
\[3k+3j+4\]
\[6k+3j+6\]
Hence either  
\[p_2\]
  or  
\[p_3\]
  is divisible by 3, contradicting that they are prime, so  
\[d\]
  is a multiple of 2 and 3 i.e. 6.
In general for a sequence of  
\[m\]
  primes, the common difference must be a multiple of  
\[m!\]
.