# The sum of the first 12 terms of an arithmetic series is 186, and the 20th term is 83. What is the sum of the first 40 terms?

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Noah G Share
Mar 8, 2018

${s}_{40} = 3420$

#### Explanation:

We know that ${s}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$

Thus

$186 = \frac{12}{2} \left(2 a + \left(11\right) d\right)$

$186 = 6 \left(2 a + 11 d\right)$

$31 = 2 a + 11 d$

Now we know that ${t}_{n} = a + \left(n - 1\right) d$.

$83 = a + \left(20 - 1\right) d$

$83 = a + 19 d$

We now have a system of equations:

$\left\{\begin{matrix}31 = 2 a + 11 d \\ 83 = a + 19 d\end{matrix}\right.$

Substituting (2) into (1), we get

$31 = 2 \left(83 - 19 d\right) + 11 d$

$31 = 166 - 38 d + 11 d$

$- 135 = - 27 d$

$d = 5$

Now solving for $a$:

$83 - 19 \left(5\right) = - 12$

The sum is once again given by

${s}_{40} = \frac{40}{2} \left(2 \left(- 12\right) + \left(39\right) 5\right)$

${s}_{40} = 20 \left(171\right)$

${s}_{40} = 3420$

Hopefully this helps!

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