An arithmetic sequence takes the form a, a+d, a+2d, a+3d, etc. The formula for the nth term (n starting at 0) where t0=a, t1=a+d, etc. is tn=a+nd. In this problem t0=5 and d=6, so tn=5+6n. If n starts at 1 then t1=a and tn=a+(n-1)d.
There's also a formula for the sum of the terms. Since each term contributes 5 to the sum we have to add as many 5s as there are terms. So part of the sum is going to be 5n where n is the actual number of terms. For just 1 term (which would be t1=5) the sum is obviously 5, because n-1=0, and the sum s1=t1. What is t2? s2=t1+t2=5+(5+6)=2×5+6=16.
t3=5+(5+6)+(5+2×6)=3×5+6(1+2)=15+18=33=5+11+17.
Now we can see that we have sn=5n+6(1+2+3+...+(n-1)).
The formula for the sum of the first n natural numbers is ½n(n-1) (this is easily provable). When n=3 this becomes ½(3)(2)=3. And this is the correct multiple of the difference d=6, so the formula for the sum of the terms in the arithmetic sequence is:
sn=5n+3n(n-1). (Note that 3=½×6.)
But this formula can be simplified: sn=5n+3n2-3n=3n2+2n=n(3n+2).
Check this by putting n=3: sum of the first three terms s3=3×11=33 which is what's expected.
The question doesn't say which formula is required but you have two formulas, one for the nth term (n starts at 1) and the other for the sum of the terms.