# What is the sum of the arithmetic sequence 140, 134, 128…, if there are 33 terms?

Apr 10, 2016

We first work out the 33th term, the general formula for a sequence like this being ${a}_{n} = {a}_{0} + n \times d$, $d$ being the difference.

#### Explanation:

Every term is the previous plus or minus the differnce $d = - 6$. So to get to the 33th term we take 32 of these steps (remember we call the first term ${a}_{0}$ and the 33th is thus ${a}_{32}$ -- remember this!!):

${a}_{32} = 140 + 32 \times \left(- 6\right) = - 52$

The sum is $33 \times$ the average, or

$\sum = 33 \times \frac{140 + \left(- 52\right)}{2} = 33 \times \frac{88}{2} = 1452$