# How do you find the sum of the arithmetic series -12-9-6-...+39?

May 9, 2016

There are 2 formulae to do this, but you need to find how many terms there are first.

"S_18 = 243

#### Explanation:

The sum of an Arithmetic series can be found from:

${S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right] \text{ }$ or $\text{ } {S}_{n} = \frac{n}{2} \left(a + l\right)$

However both of these require knowing how many terms there are.

Each term is defined by ${T}_{n} = a + \left(n - 1\right) d$

$a = - 12 , \text{ " d = 3, " } {T}_{n} = 39$

${T}_{n} = \left(- 12\right) + \left(n - 1\right) \left(3\right) = 39 \text{ solve to find n}$

$- 12 + 3 n - 3 = 39$
$3 n = 39 + 15$
$3 n = 54$
$n = 18 \text{ there are 18 terms}$

Now we can use either formula for the sum of the first 18 terms.
${S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right] \text{ }$ or $\text{ } {S}_{n} = \frac{n}{2} \left(a + l\right)$

S_18 = 18/2[2(-12) + (18-1)3] " or $\text{ } {S}_{18} = \frac{18}{2} \left(- 12 + 39\right)$

${S}_{18} = 9 \left[- 24 + 51\right] \text{ }$ or $\text{ } {S}_{18} = 9 \left(27\right)$
${S}_{18} = 9 \times 27 \text{ }$ or $\text{ } {S}_{18} = 243$
"S_18 = 243

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