The first term of the arithmetic progression is given as \(a = 6\).
The fifth term is given by the formula \(a + 4d = 12\).
Substituting \(a = 6\), we have:
\(6 + 4d = 12\)
\(4d = 6\)
\(d = 1.5\)
The sum of the first \(n\) terms of an arithmetic progression is given by:
\(S_n = \frac{n}{2} (2a + (n-1)d)\)
We know \(S_n = 90\), \(a = 6\), and \(d = 1.5\), so:
\(\frac{n}{2} (2 \times 6 + (n-1) \times 1.5) = 90\)
\(\frac{n}{2} (12 + 1.5n - 1.5) = 90\)
\(\frac{n}{2} (10.5 + 1.5n) = 90\)
\(n(10.5 + 1.5n) = 180\)
\(1.5n^2 + 10.5n - 180 = 0\)
Divide the entire equation by 1.5:
\(n^2 + 7n - 120 = 0\)
Solving this quadratic equation using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 7\), \(c = -120\):
\(n = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-120)}}{2 \times 1}\)
\(n = \frac{-7 \pm \sqrt{49 + 480}}{2}\)
\(n = \frac{-7 \pm \sqrt{529}}{2}\)
\(n = \frac{-7 \pm 23}{2}\)
\(n = 8\) or \(n = -15\)
Since \(n\) must be positive, \(n = 8\).
