an arithmetic series has un =72-6n if the sum of first n terms of the series is 378 ; find n

To find \( n \) for which the sum of the first \( n \) terms of the arithmetic series \( u_n = 72 - 6n \) equals 378, we can follow these steps:

1. **Find the first term (\( a \)) and the common difference (\( d \))**:
   - The first term \( a \) is found by substituting \( n = 1 \) into the general term formula \( u_n \):
     \[
     a = 72 - 6 \times 1 = 72 - 6 = 66
     \]
   - The second term \( u_2 \) is:
     \[
     u_2 = 72 - 6 \times 2 = 72 - 12 = 60
     \]
   - The common difference \( d \) is:
     \[
     d = u_2 - u_1 = 60 - 66 = -6
     \]

2. **Sum of the first \( n \) terms (\( S_n \))**:
   The formula for the sum of the first \( n \) terms of an arithmetic series is:
   \[
   S_n = \frac{n}{2} (2a + (n-1)d)
   \]
   Substituting the known values \( a = 66 \) and \( d = -6 \) into the formula:
   \[
   S_n = \frac{n}{2} \left[2(66) + (n-1)(-6)\right]
   \]
   Simplify inside the brackets:
   \[
   S_n = \frac{n}{2} \left[132 - 6(n-1)\right]
   \]
   \[
   S_n = \frac{n}{2} \left[132 - 6n + 6\right]
   \]
   \[
   S_n = \frac{n}{2} \left[138 - 6n\right]
   \]
   \[
   S_n = \frac{n}{2} \left[138 - 6n\right] = 69n - 3n^2
   \]

3. **Set the sum \( S_n \) equal to 378 and solve for \( n \)**:
   \[
   69n - 3n^2 = 378
   \]
   Rearrange to form a quadratic equation:
   \[
   3n^2 - 69n + 378 = 0
   \]
   Simplify by dividing the entire equation by 3:
   \[
   n^2 - 23n + 126 = 0
   \]

4. **Solve the quadratic equation**:
   Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -23 \), and \( c = 126 \):
   \[
   n = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 1 \cdot 126}}{2 \cdot 1}
   \]
   \[
   n = \frac{23 \pm \sqrt{529 - 504}}{2}
   \]
   \[
   n = \frac{23 \pm \sqrt{25}}{2}
   \]
   \[
   n = \frac{23 \pm 5}{2}
   \]

   This gives two solutions:
   \[
   n = \frac{23 + 5}{2} = 14 \quad \text{and} \quad n = \frac{23 - 5}{2} = 9
   \]

Since \( n \) must be a positive integer, we have two valid solutions: \( n = 14 \) and \( n = 9 \).

Would you like further details or have any questions? Here are some other questions you might consider:

1. How do we determine the nth term of an arithmetic series?
2. Can you show me more examples of solving arithmetic series problems?
3. What is the difference between arithmetic and geometric series?
4. How do you derive the sum formula for an arithmetic series?
5. What are some practical applications of arithmetic series?

**Tip**: Always check your solutions by substituting back into the original equations to verify correctness.

Discover how to find n in an arithmetic series where the sum of the first n terms is 378. Follow a detailed step-by-step solution using the formula for the sum of arithmetic series and the quadratic formula. Suitable for high school students.

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