An ordinary annuity such as this has 4 components.
1) K: the payment amount 2) n: the number of years for which payments are made 3) i: the effective rate of interest for any given year 4) The time at which the annuity is valued (in this case at time n)
The expression of value at the end of n years is:
[; K(1+i)^n + K(1+i)^(n-1) + \cdots + K(1+i)^2 + K(1+i) + K ;] .
To see how this expression (and the equation) comes about, think about each payment as a separate investment.
The first payment, made at t=0 grows to [; K(1+i)n ;] over n years.
The second payment, made at t=1 grows to [; K(1+i)n-1 ;] over n-1 years
Continuing, we see that we make a total of n+1 payments of K at times [; 0, 1, 2, 3, \ldots, n ;] where the [; jth ;] payment grows over a period of time equal to [; n-j ;] years.
Summing over all of these investments gives you the amount that the annuity grows to at the end of year n.
Given an amount that this annuity grows to, such as FV, we can write the equation of value:
[; FV = K(1+i)^n + K(1+i)^(n-1) + \cdots + K(1+i)^2 + K(1+i) + K ;] .
In your problem, we have
K=1000 n=5 FV=5879.09
which gives the equation:
[; 5879.09 = 1000(1+i)^5 + 1000(1+i)^4 + \cdots + 1000(1+i)^2 + 1000(1+i) + 1000 ;]
[; 5.87909 = (1+i)^5 + (1+i)^4 + (1+i)^3 + (1+i)^2 + (1+i) + 1 ;]
To find a solution i, we write this as a quintic polynomial in i
[; P(i) = (1+i)^5 + (1+i)^4 + (1+i)^3 + (1+i)^2 + (1+i) -4.87909 ;] ,
graph the polynomial in our favorite graphing utility, and use one of its functions to find one of the five solutions to the polynomial that makes sense in this situation.