How do you solve this problem using sums of arithmetic and geometric sequences.
"On the first day of each year, the Ortegas invest $1000 at 6% interest compounded annually. Find the value of their investment in 10 years."
How do you do that in one formula?
An annuity is a series of payments made at equal intervals of time. There are many types of annuities.
Contingent annuities depend on uncertain future events for their number of payment periods. Examples are pensions which typically stop at the death of the pensioner and life insurance policies which guarantee payments for the insureds lifetime.
Annuities certain guarantee a number of payment periods and the length of these periods.
Perpetuities continue forever, the number of payments being unlimited while the length of the payment periods are definite.
Annuities certain come in two varieties, ordinary annuities and annuities due. Ordinary annuities make payments at the end of each payment period while annuities due make payments at the beginning of each payment period.
Deferred annuities are annuities certain where the payments do not start for a pre-stipulated number of payment periods.
While dererred annuities are typically ordinary annuities, they are sometimes annuities due.
Simple annuities have the interest bearing periods the same as the payment periods.
General annuities have the interest bearing periods different from the payment periods.
Ordinary Annuity
What will an amount of R dollars deposited in a savings bank at the end of each year for N years, and earning I% compounded annually, amount to in N years.
Stated another way, what will R dollars deposited in a bank account at the end of each of n periods per year, and earning interest at I/n, compounded n times per year, amount to in N years?
Remember, an ordinary annuity consists of a definite number of deposits made at the ENDS of equal intervals of time. An annuity due consists of a definite number of deposits made at the BEGINNING of equal intervals of time.
For an ordinary annuity over n payment periods, n deposits are made at the end of each period but interest is paid only on (n - 1) of the payments, the last deposit drawing no interest, obviously. In the annuity due, over the same n periods, interest accrues on all n payments but there is no payment made at the end of the nth period. Consider the deposit of $R at the end of each year for 5 years at an iterest rate of I% compounded annually.
The compound interest on the first deposit at the end of the first year is S1 = R(1 + i)^(n-1).
The compound interest on the second deposit at the end of the second year is S2 = R(1+i)^(n-2).
The compound interest on the third deposit at the end of the third year is S3 = R(1+i)^(n-3)
The compound interest on the fourth deposit at the end of the fourth year is S4 = R(1+i).
There is no interest gained on the fifth deposit at the end of the fifth year.
The final accumulation is S = R(1+i)^(n-1) + R(1+i)^(n-2) + R(1+i)^(n-3) + R(1+i) + R.
Reversing this, S = R + R(1+i) + R(1+i)^(n-3) + R(1+i)^(n-2) + R(1+i)^(n-1)
Note that this is a geometric progression with first term a = R, common factor r = (1+i) and the number of terms n = n.
As we know, the sum of the terms of such a progression derives from
………………..S = a(r^n - 1)/((r - 1).
Therefore, the formula for determining the accumulation of a series of periodic deposits, made at the end of each period, over a given time span, an ordinary annuity, becomes
………………..S(n) = R[(1+i)^n - 1]/(1+i-1) = R[(1+i)^n - 1]/i
where S(n) = the accumulation over the period of n intervals, R = the periodic deposit, n = the number of interest paying periods, and i = the annual interest % divided by 100 divided by the number of interest paying periods per year.
When an annuity is computed on the basis of the payments being made at the beginning of each period, an annuity due, the total accumulation is based on one more period minus the last payment. Thus, the total accumulation becomes
.....................S(n+1) = R[(1 + i)^(n+1) - 1]/i - R
Example: $200 deposited annually for 5 years at 12% annual interest compounded annually. Therefore, R = 200, n = 5, and i = .12.
Ordinary Annuity
................................Deposit........Interest.........Balance
Beginning of month.. 1........0................0...................0
End of month..........1......200...............0.................200
Beg. of month.........2.......0.................0................200
End of month..........2......200..............24................424
Beg. of month.........3.......0.................0................424
End of month..........3......200............50.88...........674.88
Beg. of month.........4.......0.................0.............674.88
End of month..........4.....200.............80.98...........955.86
Beg. of month.........5.......0.................0.............955.86
End of month..........5.....200............114.70.........1270.56
............................S = R[(1 + i)^n - 1]/i = 200[(1.12)^5 - 1]/.12 = $1270.56
Annuity Due
..................................Deposit.......Interest..........Balance
Beginning of month 1......200..............0..................200
End of month..........1.......0...............24.................224
Beg. of month.........2.....200..............0..................424
End of month..........2.......0.............50.88...........474.88
Beg. of month.........3.....200..............0...............674.88
End of month..........3.......0.............80.98...........755.86
Beg. of month.........4.....200..............0...............955.86
End of month..........4.......0...........114.70..........1070.56
Beg. of month.........5.....200..............0..............1270.56
End of month..........5.......0...........152.47..........1423.03
............................Sn = [R[(1 + i)^(n +1) - 1]/i - R] = 200[(1.12)^6 - 1]/.12 - 200 = $1,423.03
In your case, R = the periodic deposit = $1000, n = the number of interest paying periods = 6, and i = the annual interest % divided by 100 divided by the number of interest paying periods per year = 06.
I'll let you do the math.
