Question 195291
Take note that the first term is 14 and we're adding 3 to each term to get the next term. So this tells us that {{{a[0]=14}}} and {{{d=3}}}



So the general formula {{{a[n]=dn+a[0]}}} (which is an arithmetic sequence) becomes


{{{a[n]=3n+14}}} where "n" starts at 0



Note: if you want to start at n=1, then simply subtract 3 from 14 (to scale back the terms) to get {{{a[n]=3n+11}}} where "n" starts at 1.