Find the sum of the arithmetic sequence. 3,5,7,9, ...,21

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