Let Present investments = P
Assume constant interest, i=4% (annual)
period, n = 10 (years)
Annual withdrawal, A = 7500
Use amortization formula, (opposite of borrowing)
[tex]P=\frac{A((1+i)^n-1)}{i(1+i)^n}[/tex]
[tex]=\frac{7500*((1+0.04)^10-1)}{0.04*(1+0.04)^10}[/tex]
[tex]=$60831.72[/tex] to the nearest cent
Answer: $60831.72 will be needed to maintain $7500 scholarship for 10 years
