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# 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?

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!