Find the monthly house payments necessary to amortize a \( 10.8 \% \) loan of \( \$ 191,200 \) over 20 years. The payment size is \( \$ \square \). (Round to the nearest cent.)

To find the monthly house payments necessary to amortize a loan of $191,200 with an interest rate of 10.8% over 20 years, we can use the formula for monthly payments: \[ M = \frac{P \times r \times (1 + r)^n}{(1 + r)^n - 1} \] Where: - \( M \) is the monthly payment - \( P \) is the principal amount (initial loan amount) - \( r \) is the monthly interest rate (annual interest rate divided by 12) - \( n \) is the number of payments (20 years * 12 months/year) Given: - Principal amount \( P = \$191,200 \) - Annual interest rate \( r = 10.8\% \) - Number of payments \( n = 20 \times 12 = 240 \) months Let's calculate the monthly payment using the formula. Calculate the value by following steps: - step0: Calculate: \(\frac{191200\times 0.108\left(1+0.108\right)^{240}}{\left(\left(1+0.108\right)^{240}-1\right)}\) - step1: Remove the parentheses: \(\frac{191200\times 0.108\left(1+0.108\right)^{240}}{\left(1+0.108\right)^{240}-1}\) - step2: Add the numbers: \(\frac{191200\times 0.108\times 1.108^{240}}{\left(1+0.108\right)^{240}-1}\) - step3: Add the numbers: \(\frac{191200\times 0.108\times 1.108^{240}}{1.108^{240}-1}\) - step4: Convert the expressions: \(\frac{191200\times 0.108\left(\frac{277}{250}\right)^{240}}{1.108^{240}-1}\) - step5: Convert the expressions: \(\frac{191200\times 0.108\left(\frac{277}{250}\right)^{240}}{\left(\frac{277}{250}\right)^{240}-1}\) - step6: Multiply: \(\frac{\frac{6453\times 277^{240}}{5^{721}\times 2^{236}}}{\left(\frac{277}{250}\right)^{240}-1}\) - step7: Subtract the numbers: \(\frac{\frac{6453\times 277^{240}}{5^{721}\times 2^{236}}}{\frac{277^{240}-250^{240}}{250^{240}}}\) - step8: Multiply by the reciprocal: \(\frac{6453\times 277^{240}}{5^{721}\times 2^{236}}\times \frac{250^{240}}{277^{240}-250^{240}}\) - step9: Rewrite the expression: \(\frac{6453\times 277^{240}}{5^{721}\times 2^{236}}\times \frac{125^{240}\times 2^{240}}{277^{240}-250^{240}}\) - step10: Rewrite the expression: \(\frac{6453\times 277^{240}}{5^{721}\times 2^{236}}\times \frac{5^{720}\times 2^{240}}{277^{240}-250^{240}}\) - step11: Reduce the numbers: \(\frac{6453\times 277^{240}}{5}\times \frac{2^{4}}{277^{240}-250^{240}}\) - step12: Multiply the fractions: \(\frac{6453\times 277^{240}\times 2^{4}}{5\left(277^{240}-250^{240}\right)}\) - step13: Multiply: \(\frac{103248\times 277^{240}}{5\times 277^{240}-5\times 250^{240}}\) The monthly house payment necessary to amortize a loan of $191,200 with an interest rate of 10.8% over 20 years is approximately $20,649.60. Therefore, the payment size is $20,649.60.