# The sum of 40 numbers in an arithmetic sequence is 8000. If the first number is the sequence is 83, what is the last number?

Mar 8, 2016

$317$

#### Explanation:

In an arithmetic sequence the difference between consecutive members is constant. Therefore the sequence is
$83 , 83 + d , 83 + 2 d , \ldots . .83 + 39 d$

The $i$th member is $83 + \left(i - 1\right) d$

As shown on Arithmetic Sequences the formula for the sum of members is
${\sum}_{0}^{n - 1} \left(a + k d\right) = \frac{n}{2} \cdot \left(2 a + \left(n - 1\right) d\right)$

In this case
${\sum}_{0}^{39} = \frac{40}{2} \cdot \left(2 \cdot 83 + 39 \cdot d\right) = 8000$
$166 + 39 d = 400$
$39 d = 234$

$d = \frac{234}{39} = 6$

The $40$th number is therefore $83 + 39 \cdot 6 = 317$