This document discusses general annuities where the payment interval is different from the compounding period. It provides examples of general annuities including monthly car payments with annual interest compounding and semi-annual debt payments with monthly compounding. The document also presents the formulas for calculating the future and present value of a general ordinary annuity. It includes two examples of using the formulas to solve problems involving equivalent interest rates, payment amounts, and time periods.

1. GENERAL ANNUITY
ALAN S. ABERILLA

2. GENERAL ANNUITY – an annuity where the length of the payment interval is not the same as the
length of the interest compounding period
GENERAL ORDINARY ANNUITY – a general annuity in which the periodic payment is made at the end
of the payment interval
Examples of General Annuity:
1. Monthly is installment payment of a car, lot, or house with an interest rate that is compounded
annually.
2. Paying a debt semi-annually when the interest is compounded monthly.

3. Future and Present Value of a General Ordinary Annuity
The future value F and present value P of a general ordinary annuity is given by
(1 + j)n – 1 (1 + j)n - 1
F= R and P = R
j j
where R is the regular paymend
j is the equivalent interest rate per payment interval converted from the interest
rate per
period
n is the number of payments
Note: When solving for an equivalent rate, sa.y j, six or more decimal places will be used

4. Example 1. Cris started to deposit P 1,000.00 in a fund that pays 6% compounded quarterly. How much will be in
the fund after 15 years?
Given: R = P 1,000.00 n = (12)(15) = 180 payments i(4) = 0.06 = 4
(1) Convert 6% compounded quarterly to its equivalent (2) Apply the formula in finding the future value of an
interest rate for monthly payment interval. ordinary annuity using the value computed equivalent
rate.
F1 = F2
P{(1 + i(12))/12)} (12)t = P{(1 + i(4))/4)} (4)t
{(1 + i(12))/12)} (12) = {(1 + 0.06/4)} (4)
{(1 + i(12))/12)} (12) = (1.015)4
{(1 + i(12))/12)} = {(1.015)4} (1/12)
i(12))/12 = (1.015)(1/3) – 1
i(12))/12 = 0.00497521 = j
F = R [(1 + j)n – 1)/j)]
F = P 1,000.00[(1 + 0.00497521)180 – 1)/0.00497521]
F = P 290,082.51
Thus, Cris will have P 290,082.51 in the fund after 20
years.

5. Example 2. A teacher saves P 5,000.00 every 6 months in a bank that pays 0.25% compounded monthly. How
much will be her savings after 10 years?
Given: R = P 5,000.00 n = (2)(10) = 20 payments i(12) = 0.25% = 0.0025 m = 12
(1) Convert 0.25% compounded monthly to its equivalent (2) Apply the formula in finding the future value of
an
interest rate for each semi-annual payment interval. ordinary annuity using the value computed
equivalent rate.F1 = F2
P{(1 + i(2))/2)} (2)t = P{(1 + i(12))/12)}(12)t
{(1 + i(2))/2)} (2) = {(1 + 0.0025/12)} (12)
{(1 + i(2))/2)} (2) = (1.00020833)12
{(1 + i(2))/2)} = {(1.015)12} (1/2)
i(2))/2 = (1.00020833)(6) – 1
i(2))/2 = 0.00125063 = j
F = R [(1 + j)n – 1)/j)]
F = P 5,000.00[(1 + 0.00125063)20 – 1)/0.00125063]
F = P 101,197.06
Thus, the teacher will be able to save P 101,197.06 after
10 years.

6. ACTIVITY 8
Solve the following:
1. Alfred borrowed an amount of money from Helen. He agrees to pay the principal plus interest by paying
P 38,973.76 each year for 3 years. How much money did he borrow if interest is 8% compounded
quarterly?
2. Mrs. Torres would like to buy a television (TV) set payable for 6 months starting at the end of the month.
How much is the cost of the TV set if her monthly payment is P 3,000.00 and interest is 9% compounded
semi-annually?
3. ABC bank pays interest rate at the rate of 2% compounded quarterly. How much will Ken have in the
bank at the end of 5 years if he deposits P 3,000.00 every month?
4. A sala set is for sale at P 16,000.00 in cash or on monthly installment of P 2,950.00 for 6 months at 12%
compounded semi-annually. Which is lower: the cash price or the present value of the installment term?