Part 2 General Annuities

4. Caryn has enough money in her savings account withdraw $850 at the beginning of each year for 10 years, beginning 3 years from now. If money earns 8% compounded quarterly, how much does Caryn have now?

5. What amount would be required quarterly to amortize a debt of $45,000 in 10 years, if the interest rate is 9% compounded monthly?

6. If money deposits of $15 a month earn interest at 12% compounded quarterly, how long will it take to save $5,000 if the deposits are made:

a. At the beginning of each month?

b. At the end of each month?

Financial Mathematics

FORMULA SHEET

i = j / m

I = Prt

t = I / Pr

P = I / rt

S = P(1 + i)ⁿ

f = (1 + i)^m - 1

n = ln (S / P)

ln (1 + i)

Sn = R[(1 + p)ⁿ - 1]

p

R = Sn

[(1 + p)ⁿ - 1] / p

14. = ln [1 + pSn/R] ln (1 + p)

Sn(due) = R[(1 + p)ⁿ - 1](1 + p)

p

n = ln [1 + [pSn(due) / R(1 + p)] ln(1 + p)

14. = -ln[1 - (p[1 + p]^dAn(def))/R] ln(1 + p)

An(def) = R [1 - (1 + p)^-n] p(1 + p)^d

A = R / p

m = j / i

S = P(1 + rt)

r = I / Pt

P = S / (1 + rt) = S(1 + i)^-n

c = # of compoundings/# of payments

p = (1 + i)^c - 1

i = [S / P] ^1/n - 1

An = R[1 - (1 + p)^-n]

p

R = An

[1 - (1 + p)^-n] / p

14. = -ln [1 - pAn/R] ln (1 + p)

An(due) = R[1 - (1 + p)^-n](1 + p)

p

n = -ln[1 - [pAn(due) / R(1 + p)] ln(1 + p)

d = -ln{R[1-(1 + p)^-n] / pAn(def)} ln(1 + p)

Sn(def) = Sn

A(due) = (R / p)(1 + p)