The sum of any arithmetic sequence is the average of the first and last terms times the number of terms.
Any term in an arithmetic sequence is:
a(n)=a+d(n-1), where a=initial term, d=common difference, n=term number
So the first term is a, and the last term is a+d(n-1) so the sum can be expressed as:
s(n)(a+a+d(n-1))(n/2)
s(n)=(2a+dn-d)(n/2)
s(n)=(2an+dn^2-dn)/2
However we need to know how many terms are in the sequence.
a(n)=a+d(n-1), and we know a=3 and d=2 and a(n)=21 so
21=3+2(n-1)
18=2(n-1)
9=n-1
10=n so there are 10 terms in the sequence.
s(n)=(2an+dn^2-dn)/2, becomes, a=3, d=2, n=10
s(10)=(2*3*10+2*10^2-2*10)/2
s(10)=(60+200-20)/2
s(10)=240/2
s(10)=120
