# What is the sum of a 22-term arithmetic sequence where the first term is 56 and the last term is -28?

Apr 10, 2016

308

#### Explanation:

The sequence is {56+(n-1)d}, 2 = 1, 2, 3, ...,22, wher d is the common difference.

The last term (n=22) is $- 28 = 56 + 21 d$,
So, $d = - \frac{82}{21} = - 4$.

The sum = $\left(22\right) \left(56\right) + \left(d + 2 d + 3 d , \ldots + 21 d\right) = 1232 - \left(21\right) \left(\frac{22}{2}\right) \left(- 4\right)$. using $1 + 2 + 3. . . + n = \frac{n \left(n + 1\right)}{2}$
The sum simplifies to 308.