Problems involving the sum of arithmetic sequences with given first terms, common differences, and number of terms.

Problem 1: Find the sum of the first 10 terms of the arithmetic sequence with \( a_1 = 3 \) and \( d = 2 \).

#### Solution:
The formula for the sum of the first \( n \) terms of an arithmetic sequence is:
\[
S_n = \frac{n}{2} \left( 2a_1 + (n-1)d \right)
\]
Here, \( a_1 = 3 \), \( d = 2 \), and \( n = 10 \).

Substitute the values into the formula:
\[
S_{10} = \frac{10}{2} \left( 2 \cdot 3 + (10-1) \cdot 2 \right)
\]
Simplify inside the parentheses:
\[
S_{10} = 5 \left( 6 + 9 \cdot 2 \right) = 5 \left( 6 + 18 \right) = 5 \cdot 24 = 120
\]

Thus, the sum of the first 10 terms is:
\[
\boxed{120}
\]

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Problem 5: Find the sum of the first 40 terms of the arithmetic sequence \(-7, -5, -3, -1, \ldots\).

#### Solution:
The given sequence is \(-7, -5, -3, -1, \ldots\). Here, the first term \( a_1 = -7 \) and the common difference \( d = 2 \).

Using the sum formula for an arithmetic sequence:
\[
S_n = \frac{n}{2} \left( 2a_1 + (n-1)d \right)
\]
Here, \( a_1 = -7 \), \( d = 2 \), and \( n = 40 \).

Substitute the values into the formula:
\[
S_{40} = \frac{40}{2} \left( 2 \cdot (-7) + (40-1) \cdot 2 \right)
\]
Simplify inside the parentheses:
\[
S_{40} = 20 \left( -14 + 39 \cdot 2 \right) = 20 \left( -14 + 78 \right) = 20 \cdot 64 = 1280
\]

Thus, the sum of the first 40 terms is:
\[
\boxed{1280}
\]

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Problem 9: Find the sum \( 1 + 7 + 13 + \cdots + 145 \).

#### Solution:
The given sequence is \( 1, 7, 13, \ldots, 145 \). This is an arithmetic sequence where the first term \( a_1 = 1 \) and the common difference \( d = 6 \).

First, we need to find the number of terms \( n \) in the sequence. The \( n \)-th term of an arithmetic sequence is given by:
\[
a_n = a_1 + (n-1)d
\]
Here, \( a_n = 145 \), \( a_1 = 1 \), and \( d = 6 \). Substitute these values:
\[
145 = 1 + (n-1) \cdot 6
\]
Solve for \( n \):
\[
145 = 1 + 6(n-1) \implies 145 = 1 + 6n - 6 \implies 145 = 6n - 5 \implies 150 = 6n \implies n = 25
\]

Now, we know there are 25 terms in the sequence. Use the sum formula:
\[
S_n = \frac{n}{2} \left( a_1 + a_n \right)
\]
Here, \( n = 25 \), \( a_1 = 1 \), and \( a_n = 145 \). Substitute these values:
\[
S_{25} = \frac{25}{2} \left( 1 + 145 \right) = \frac{25}{2} \cdot 146 = 25 \cdot 73 = 1825
\]

Thus, the sum of the sequence is:
\[
\boxed{1825}
\]

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Problem 12: An arithmetic sequence has \( a_1 = 4 \) and \( a_7 + a_8 = -57 \). Find the sum of its first 80 terms.

#### Solution:
The given sequence has \( a_1 = 4 \). Let the common difference be \( d \).

The \( n \)-th term of an arithmetic sequence is:
\[
a_n = a_1 + (n-1)d
\]
For \( a_7 \):
\[
a_7 = a_1 + 6d = 4 + 6d
\]
For \( a_8 \):
\[
a_8 = a_1 + 7d = 4 + 7d
\]
We are given that \( a_7 + a_8 = -57 \). Substitute the expressions for \( a_7 \) and \( a_8 \):
\[
(4 + 6d) + (4 + 7d) = -57
\]
Simplify:
\[
8 + 13d = -57 \implies 13d = -65 \implies d = -5
\]

Now, we need to find the sum of the first 80 terms. Using the sum formula:
\[
S_n = \frac{n}{2} \left( 2a_1 + (n-1)d \right)
\]
Here, \( n = 80 \), \( a_1 = 4 \), and \( d = -5 \). Substitute these values:
\[
S_{80} = \frac{80}{2} \left( 2 \cdot 4 + (80-1)(-5) \right)
\]
Simplify inside the parentheses:
\[
S_{80} = 40 \left( 8 + 79 \cdot (-5) \right) = 40 \left( 8 - 395 \right) = 40 \cdot (-387) = -15480
\]

Thus, the sum of the first 80 terms is:
\[
\boxed{-15480}
\]

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Final Answers:

1. \(\boxed{120}\)
5. \(\boxed{1280}\)
9. \(\boxed{1825}\)
12. \(\boxed{-15480}\)