Question

Find the amount of a 10 year mortgage at 9.9% interest compounded monthly where the monthly payment is $448.21. the amount of the mortgage is $

Answer 1

`\bf ~~~~~~~~~~~~ \textit{Amortized Loan Value} \\\\ pymt=P\left[ \cfrac{(r)/(n)}{1-\left( 1+ (r)/(n)\right)^(-nt)} \right]`

`\bf ~~~~~~ \begin{cases} P= \begin{array}{llll} \textit{original amount deposited}\\ \end{array} & \begin{array}{llll} \end{array}\\ pymt=\textit{periodic payments}\dotfill &448.21\\ r=rate\to 9.9\%\to (9.9)/(100)\dotfill &0.099\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &10 \end{cases}`

`\bf 448.21=P\left[ \cfrac{(0.099)/(12)}{1-\left( 1+ (0.099)/(12)\right)^(-12\cdot 10)} \right]\implies 448.21=P\left[ \cfrac{0.00825}{1-\left( 1.00825\right)^(-120)} \right] \\\\\\ 448.21\approx P \left[ \cfrac{0.00825}{0.6269111} \right]\implies 448.21\approx P(0.0131598) \\\\\\ \cfrac{448.21}{0.0131598}\approx P\implies 34059.02825 \approx P`