\[a_n=11n-4\]
. 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.