Respuesta :
Answer:
[tex]x = 97[/tex]
Step-by-step explanation:
Given
[tex]t = 20[/tex] --- time (years)
[tex]A =1000[/tex] --- amount
[tex]r = 10\%[/tex] --- rate of interest
Required
The last 10 payments (x)
First, calculate the end of year 1 payment
[tex]y_1(end) = 10\% * 1000 * 150\%[/tex]
[tex]y_1(end) = 150[/tex]
Amount at end of year 1
[tex]A_1=A - y_1(end) - r * A[/tex]
[tex]A_1=1000 - (150 - 10\% * 1000)[/tex]
[tex]A_1 =1000 - (150- 100)[/tex]
[tex]A_1 =950[/tex]
Rewrite as:
[tex]A_1 = 0.95 * 1000^1[/tex]
Next, calculate the end of year 1 payment
[tex]y_2(end) = 10\% * 950 * 150\%[/tex]
[tex]y_2(end) = 142.5[/tex]
Amount at end of year 2
[tex]A_2=A_1 - (y_2(end) - r * A_1)[/tex]
[tex]A_2=950 - (142.5 - 10\%*950)[/tex]
[tex]A_2 = 902.5[/tex]
Rewrite as:
[tex]A_2 = 0.95 * 1000^2[/tex]
We have been able to create a pattern:
[tex]A_1 = 1000 * 0.95^1 = 950[/tex]
[tex]A_2 = 1000 * 0.95^2 = 902.5[/tex]
So, the payment till the end of the 10th year is:
[tex]A_{10} = 1000*0.95^{10}[/tex]
[tex]A_{10} = 598.74[/tex]
To calculate X (the last 10 payments), we make use of the following geometric series:
[tex]Amount = \sum\limits^{9}_{n=0} x * (1 + r)^n[/tex]
[tex]Amount = \sum\limits^{9}_{n=0} x * (1 + 10\%)^n[/tex]
[tex]Amount = \sum\limits^{9}_{n=0} x * (1 + 0.10)^n[/tex]
[tex]Amount = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]
The amount to be paid is:
[tex]Amount = A_{10}*(1 + r)^{10}[/tex] --- i.e. amount at the end of the 10th year * rate of 10 years
[tex]Amount = 1000 * 0.95^{10} * (1+r)^{10}[/tex]
So, we have:
[tex]Amount = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]
[tex]\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+r)^{10}[/tex]
[tex]\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+10\%)^{10}[/tex]
[tex]\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+0.10)^{10}[/tex]
[tex]\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1.10)^{10}[/tex]
The geometric sum can be rewritten using the following formula:
[tex]S_n = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]
[tex]S_n =\frac{a(r^n - 1)}{r -1}[/tex]
In this case:
[tex]a = x[/tex]
[tex]r = 1.10[/tex]
[tex]n =10[/tex]
So, we have:
[tex]\frac{x(r^{10} - 1)}{r -1} = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]
[tex]\frac{x((1.10)^{10} - 1)}{1.10 -1} = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]
[tex]\frac{x((1.10)^{10} - 1)}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]
[tex]x * \frac{1.10^{10} - 1}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]
So, the equation becomes:
[tex]x * \frac{1.10^{10} - 1}{0.10} = 1000 * 0.95^{10} * (1.10)^{10}[/tex]
Solve for x
[tex]x = \frac{1000 * 0.95^{10} * 1.10^{10} * 0.10}{1.10^{10} - 1}[/tex]
[tex]x = 97.44[/tex]
Approximate
[tex]x = 97[/tex]
