Chapter 12.1, Problem 83AYU

Car Loans Phil bought a car by taking out a loan for $\$18,500$ at $0.5\%$ interest per month. Phil's normal monthly payment is $\$434.47$ per month, but he decides that he can afford to pay $\$100$ extra toward the balance each month. His balance each month is given by the recursively defined sequence

$B_0=18,500B_n=1.005B_{n-1}-534.47$.

Determine Phil's balance alter making the first payment. That is. determine $B_1$.

To determine

Phil’s balance after making the first payment. That is, $B_1$ where, Phil bought a car by taking out a loan for $\$18,500$ at $0.5\%$ interest per month. Phil’s normal monthly payment is $\$434.47$ per month, but he decides that he can afford to pay $\$100$ extra towards the balance each month. His balance each month is given by the recursively defined sequence $B_0=\$18,500B_n=1.005B_{n-1}-534.47$.

Answer to Problem 83AYU

Solution:

Phil’s balance is $\text{\$18058}\text{.03}$ after making the first payment.

Explanation of Solution

Given information:

Phil bought a car by taking out a loan for $\$18,500$ at $0.5\%$ interest per month. Phil’s normal monthly payment is $\$434.47$ per month, but he decides that he can afford to pay $\$100$ extra toward the balance each month. His balance each month is given by the recursively defined sequence $B_0=\$18,500B_n=1.005B_{n-1}-534.47$.

Explanation:

Consider the recursive relation, $B_0=\$18,500B_n=1.005B_{n-1}-534.47$.

To find Phil’s balance after making the first payment hat is $B_1$.

Substitute $n=1$ in the given recursive equation,

$B_n=1.005B_{n-1}-534.47$.

$\Rightarrow B_1=1.005B_{1-1}-534.47$

$\Rightarrow B_1=1.005B_0-534.47$

Substitute $B_0=\$18,500$ in above equation.

$\Rightarrow B_1=1.005\left(18,500\right)-534.47=\text{18058}\text{.03}$.

Therefore, Phil’s balance after making the first payment is, $B_1=\$\text{18058}\text{.03}$.