Chapter: Arithmetic Progression· Mathematics Part 1 (Algebra)
Answer: 20 months
Payments form an A.P.: \(a = 30500\), \(d = -1500\), \(S_n = 325000\).
\(325000 = \dfrac{n}{2}[2 \times 30500 + (n - 1)(-1500)]\)
\(650000 = n[61000 - 1500n + 1500] = n(62500 - 1500n)\)
Divide by 500: \(1300 = n(125 - 3n)\)
\(3n^2 - 125n + 1300 = 0\)
\(n = \dfrac{125 \pm \sqrt{15625 - 15600}}{6} = \dfrac{125 \pm 5}{6}\)
\(n = 20\) or \(n = \dfrac{130}{6}\) (not a whole number, rejected).
He will clear the amount in 20 months. Check: \(t_{20} = 30500 - 19 \times 1500 = 2000\); \(S_{20} = 10(30500 + 2000) = 325000\) ✓.