# How do you find the sum of the first 14 terms of the arithmetic series 11+7+3+...?

Mar 7, 2016

$S = - 210$

#### Explanation:

There are several formula we need to keep in mind when working with arithmetic series

1) common difference $d = {a}_{n + 1} - {a}_{n}$

2) last term : ${a}_{n} = {a}_{1} + \left(n - 1\right) d$

3) Arithmetic sum: $S = \frac{n}{2} \left({a}_{1} + {a}_{n}\right)$
An alternative formula is $S = \frac{n}{2} \left(2 {a}_{1} + \left(n - 1\right) d\right)$

Now we can begin to work this problem

We have an arithmetic series begin at 11, 7, 3 ......

We know that ${a}_{1} = \textcolor{b l u e}{11}$

Let's find the following
$d = 7 - 11 = - 4$
$\textcolor{g r e e n}{n = 14}$
${a}_{14} = 11 + \left(14 - 1\right) \left(- 4\right)$

$\textcolor{red}{{a}_{14} = - 41}$

Let's substitute it into the sum formula

$S = \frac{\textcolor{g r e e n}{14}}{2} \left(\left(\textcolor{b l u e}{11}\right) + \left(\textcolor{red}{- 41}\right)\right)$

$S = 7 \left(- 30\right)$

$S = - 210$

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If we use the alternative formula, we will still get the same answer
$d = - 4 , {a}_{1} = 11 , n = 14$

$S = \frac{14}{2} \left(2 \left(11\right) + \left(14 - 1\right) \left(- 4\right)\right)$

$S = 7 \left(22 - 52\right)$

$S = 7 \left(- 30\right)$

$S = - 210$