Successive Terms of Arithmetic Sequence Divisible by 2, 3, 5 Respectively

Consider the arithmetic sequence 7, 18, 29,... generated by the rule  
\[a_n=11n-4\]
. What value of n gives successive terms that divide by 2, 3 and 5 respectively?
\[a_n=11n-4\]
  so  
\[n\]
  must be even if 2 is to divide  
\[a_n\]
. Put  
\[n=2k\]
  then  
\[a_{2k}=22k-4\]
  is divisible by 2.
\[a_{n+1}=a_{2k+1}=11(2k+1)-4=22k+7\]
  generates the sequence 29, 51, 73, 95, 117, 139, 161, 183,... with 2nd, 5th, 8th,... terms being divisible by 3, so put  
\[k=3j-1\]
.
\[a_{n+2}=a_{2k+2}=a_{2(3j-1)+2}=a_{6j}=11(6j)-4=66j-4\]
  generates the sequence 62, 128, 194, 260, 326, 392, 458, 5285,... with 4th, 8th, 12th,... terms being divisible by 5.
Then  
\[a_{6 \times 4}=a_{24}=11 \times 24-4=260, \; a_{23}=11 \times 23-4=249, \; a_{22}=11 \times 22-4=238\]
.
2, 3, 5 divide  
\[a_{22}, \; a_{23}, \; a_{24}\]
  respectively.

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