The answer is:
120
Given the arithmetic sequence
3, 5, 7, 9, ..., 21
The first term a = 3 & a common difference d = 5 - 3 = 7 - 5 = ... = 2
If there are n number of terms in the above A.P. then last term l = 21 will be the nth term given as:
l = a + (n - 1)d
21 = 3 + (n − 1)2
n = 10
Hence the sum of given arithmetic progression (A. P.) up to 10 terms is given general formula.
Sn = n/2(a + l)
S10 = 10/2(3 + 21)
= 120