Question

A 20-year loan of 1000 is repaid with payments at the end of each year. Each of the first ten payments equals 150% of the amount of interest due. Each of the last ten payments is X. The lender charges interest at an annual effective rate of 10%. Calculate X.

Answer 1

Answer:

$x = 97$

Step-by-step explanation:

Given

$t = 20$ --- time (years)

$A =1000$ --- amount

$r = 10\%$ --- rate of interest

Required

The last 10 payments (x)

First, calculate the end of year 1 payment

$y_1(end) = 10\% * 1000 * 150\%$

$y_1(end) = 150$

Amount at end of year 1

$A_1=A - y_1(end) - r * A$

$A_1=1000 - (150 - 10\% * 1000)$

$A_1 =1000 - (150- 100)$

$A_1 =950$

Rewrite as:

$A_1 = 0.95 * 1000^1$

Next, calculate the end of year 1 payment

$y_2(end) = 10\% * 950 * 150\%$

$y_2(end) = 142.5$

Amount at end of year 2

$A_2=A_1 - (y_2(end) - r * A_1)$

$A_2=950 - (142.5 - 10\%*950)$

$A_2 = 902.5$

Rewrite as:

$A_2 = 0.95 * 1000^2$

We have been able to create a pattern:

$A_1 = 1000 * 0.95^1 = 950$

$A_2 = 1000 * 0.95^2 = 902.5$

So, the payment till the end of the 10th year is:

$A_{10} = 1000*0.95^{10}$

$A_{10} = 598.74$

To calculate X (the last 10 payments), we make use of the following geometric series:

$Amount = \sum\limits^{9}_{n=0} x * (1 + r)^n$

$Amount = \sum\limits^{9}_{n=0} x * (1 + 10\%)^n$

$Amount = \sum\limits^{9}_{n=0} x * (1 + 0.10)^n$

$Amount = \sum\limits^{9}_{n=0} x * (1.10)^n$

The amount to be paid is:

$Amount = A_{10}*(1 + r)^{10}$ --- i.e. amount at the end of the 10th year * rate of 10 years

$Amount = 1000 * 0.95^{10} * (1+r)^{10}$

So, we have:

$Amount = \sum\limits^{9}_{n=0} x * (1.10)^n$

$\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+r)^{10}$

$\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+10\%)^{10}$

$\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+0.10)^{10}$

$\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1.10)^{10}$

The geometric sum can be rewritten using the following formula:

$S_n = \sum\limits^{9}_{n=0} x * (1.10)^n$

$S_n =\frac{a(r^n - 1)}{r -1}$

In this case:

$a = x$

$r = 1.10$

$n =10$

So, we have:

$\frac{x(r^{10} - 1)}{r -1} = \sum\limits^{9}_{n=0} x * (1.10)^n$

$\frac{x((1.10)^{10} - 1)}{1.10 -1} = \sum\limits^{9}_{n=0} x * (1.10)^n$

$\frac{x((1.10)^{10} - 1)}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n$

$x * \frac{1.10^{10} - 1}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n$

So, the equation becomes:

$x * \frac{1.10^{10} - 1}{0.10} = 1000 * 0.95^{10} * (1.10)^{10}$

Solve for x

$x = \frac{1000 * 0.95^{10} * 1.10^{10} * 0.10}{1.10^{10} - 1}$

$x = 97.44$

Approximate

$x = 97$