To find \( n \) given the arithmetic series where \( v_4 = 23 \), \( v_7 = 44 \), and \( S_n = 80 \):
1. Determine the Common Difference \( d \):
- The general form for an arithmetic sequence is \( v_n = a + (n-1)d \), where \( a \) is the first term and \( d \) is the common difference.
- For \( v_4 \) and \( v_7 \):
[tex]\[
v_4 = a + 3d = 23
\][/tex]
[tex]\[
v_7 = a + 6d = 44
\][/tex]
- Subtracting these equations:
[tex]\[
(a + 6d) - (a + 3d) = 44 - 23
\][/tex]
[tex]\[
3d = 21
\][/tex]
[tex]\[
d = 7
\][/tex]
2. Find the First Term \( a \):
- Using \( v_4 \):
[tex]\[
23 = a + 3d
\][/tex]
[tex]\[
23 = a + 3 \cdot 7
\][/tex]
[tex]\[
23 = a + 21
\][/tex]
[tex]\[
a = 2
\][/tex]
3. General Form of the Arithmetic Series Sum:
- The sum of the first \( n \) terms of an arithmetic series is given by:
[tex]\[
S_n = \frac{n}{2} (2a + (n-1)d)
\][/tex]
4. Substitute the Known Values:
- Given \( S_n = 80 \), \( a = 2 \), and \( d = 7 \):
[tex]\[
80 = \frac{n}{2} (2 \cdot 2 + (n-1) \cdot 7)
\][/tex]
[tex]\[
80 = \frac{n}{2} (4 + 7n - 7)
\][/tex]
[tex]\[
80 = \frac{n}{2} (7n - 3)
\][/tex]
[tex]\[
160 = n (7n - 3)
\][/tex]
[tex]\[
7n^2 - 3n - 160 = 0
\][/tex]
5. Solve the Quadratic Equation \( 7n^2 - 3n - 160 = 0 \):
- The quadratic equation can be solved using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 7 \), \( b = -3 \), and \( c = -160 \):
[tex]\[
n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 7 \cdot (-160)}}{2 \cdot 7}
\][/tex]
[tex]\[
n = \frac{3 \pm \sqrt{9 + 4480}}{14}
\][/tex]
[tex]\[
n = \frac{3 \pm \sqrt{4489}}{14}
\][/tex]
[tex]\[
n = \frac{3 \pm 67}{14}
\][/tex]
- Solving for positive \( n \):
[tex]\[
n = \frac{3 + 67}{14} = \frac{70}{14} = 5
\][/tex]
Thus, [tex]\( n = 5 \)[/tex].
