Solve for the value of n. 
1. Sn= -15, a1=12, d= -3, n=?
2. Sn=180, a1=5, d=5, n=?

To solve for the value of \( n \) in each of these arithmetic series problems, we'll use the formula for the sum of the first \( n \) terms of an arithmetic series:

\[
S_n = \frac{n}{2} \times (2a_1 + (n-1)d)
\]

### 1. Problem 1:
Given:
- \( S_n = -15 \)
- \( a_1 = 12 \)
- \( d = -3 \)

Substitute these values into the formula:

\[
-15 = \frac{n}{2} \times \left(2(12) + (n-1)(-3)\right)
\]

Simplify the equation:

\[
-15 = \frac{n}{2} \times \left(24 - 3n + 3\right)
\]
\[
-15 = \frac{n}{2} \times (27 - 3n)
\]
\[
-30 = n(27 - 3n)
\]
\[
-30 = 27n - 3n^2
\]

Rearrange the equation to form a standard quadratic equation:

\[
3n^2 - 27n - 30 = 0
\]

Divide the entire equation by 3:

\[
n^2 - 9n - 10 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -9 \), and \( c = -10 \):

\[
n = \frac{9 \pm \sqrt{(-9)^2 - 4(1)(-10)}}{2(1)}
\]
\[
n = \frac{9 \pm \sqrt{81 + 40}}{2}
\]
\[
n = \frac{9 \pm \sqrt{121}}{2}
\]
\[
n = \frac{9 \pm 11}{2}
\]

So the two possible solutions for \( n \) are:

\[
n = \frac{9 + 11}{2} = 10 \quad \text{or} \quad n = \frac{9 - 11}{2} = -1
\]

Since \( n \) must be a positive integer, we choose \( n = 10 \).

### 2. Problem 2:
Given:
- \( S_n = 180 \)
- \( a_1 = 5 \)
- \( d = 5 \)

Substitute these values into the formula:

\[
180 = \frac{n}{2} \times \left(2(5) + (n-1)(5)\right)
\]

Simplify the equation:

\[
180 = \frac{n}{2} \times \left(10 + 5n - 5\right)
\]
\[
180 = \frac{n}{2} \times (5n + 5)
\]
\[
360 = n(5n + 5)
\]
\[
360 = 5n^2 + 5n
\]

Divide the entire equation by 5:

\[
72 = n^2 + n
\]

Rearrange the equation to form a standard quadratic equation:

\[
n^2 + n - 72 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 1 \), and \( c = -72 \):

\[
n = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-72)}}{2(1)}
\]
\[
n = \frac{-1 \pm \sqrt{1 + 288}}{2}
\]
\[
n = \frac{-1 \pm \sqrt{289}}{2}
\]
\[
n = \frac{-1 \pm 17}{2}
\]

So the two possible solutions for \( n \) are:

\[
n = \frac{-1 + 17}{2} = 8 \quad \text{or} \quad n = \frac{-1 - 17}{2} = -9
\]

Since \( n \) must be a positive integer, we choose \( n = 8 \).

### Summary:
1. For the first problem, \( n = 10 \).
2. For the second problem, \( n = 8 \).

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. How would the solution change if the common difference \( d \) was positive in problem 1?
2. What if \( S_n \) was positive instead of negative in problem 1?
3. How would you find the number of terms if only the last term and the sum were given?
4. What happens to the sum formula if \( d = 0 \)?
5. Can you derive the formula for the sum of an arithmetic series starting from the basic definition?

**Tip:** Always double-check if the solutions to quadratic equations make sense in the context of the problem, especially when considering the domain of the variable (e.g., \( n \) must be a positive integer in sequences).

Learn how to find the number of terms in arithmetic series using the sum formula. Detailed step-by-step solutions for arithmetic series problems with negative and positive common differences. Suitable for advanced high school students.

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