Question

use the formula (s_n = \frac{n(a_1 + a_n)}{2}) to evaluate the following sum.

\sum_{k=1}^{12} (-4 + (k - 1) \cdot 3)

(s_{12} =)

Explanation:

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<reasoning>

Identify the parameters of the arithmetic series

\[

$$\begin{aligned} &n = 12 \\ &a_1 = -4 + (1 - 1) \cdot 3 = -4 \\ &a_{12} = -4 + (12 - 1) \cdot 3 = -4 + 33 = 29 \end{aligned}$$

\]

Apply the arithmetic series sum formula

\[

$$\begin{aligned} &S_{12} = \frac{12 \cdot (a_1 + a_{12})}{2} \\ &S_{12} = 6 \cdot (-4 + 29) \\ &S_{12} = 6 \cdot 25 = 150 \end{aligned}$$

\]
</reasoning>

<answer>
Use the formula \(S_n = \frac{n(a_1 + a_n)}{2}\) to evaluate the following sum.
\[\sum_{k=1}^{12} (-4 + (k - 1) \cdot 3)\]
\(S_{12} =\) <blank>150</blank>
</answer>

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Answer:

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<reasoning>

Identify the parameters of the arithmetic series

\[

$$\begin{aligned} &n = 12 \\ &a_1 = -4 + (1 - 1) \cdot 3 = -4 \\ &a_{12} = -4 + (12 - 1) \cdot 3 = -4 + 33 = 29 \end{aligned}$$

\]

Apply the arithmetic series sum formula

\[

$$\begin{aligned} &S_{12} = \frac{12 \cdot (a_1 + a_{12})}{2} \\ &S_{12} = 6 \cdot (-4 + 29) \\ &S_{12} = 6 \cdot 25 = 150 \end{aligned}$$

\]
</reasoning>

<answer>
Use the formula \(S_n = \frac{n(a_1 + a_n)}{2}\) to evaluate the following sum.
\[\sum_{k=1}^{12} (-4 + (k - 1) \cdot 3)\]
\(S_{12} =\) <blank>150</blank>
</answer>

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