What is the sum of the arithmetic sequence 3, 9, 15
if there are 34 terms?

Work Shown:

a = first term = 3

d = common difference = 6

S(n) = sum of the first n terms of an arithmetic sequence

S(n) = (n/2)*(2a + d(n-1))

S(34) = (34/2)*(2*3 + 6(34-1))

S(34) = 3468

Check:

3+9+15+21+27+33+39+45+51+57+63+69+75+81+87+93+99+105+111+117+123+129+135+141+147+153+159+165+171+177+183+189+195+201 = 3468

I used GeoGebra to generate the 34 terms shown above. You could do so by hand (start at 3; add 6 to each term to get the next one), but it's a tedious busywork type of problem in my opinion. It's best left to computer software.