Question

Find the monthly payment of the given loan.

A loan of 10000 dollars amortized over 48 months at 8%

Answer 1

275
10000*0.08*4=3200
10000+3200 = 13200 - total sum to payoff
13200/48=275

Answer 2

hmmmm that wasn't the correct answer

Answer 3

what is the correct answer?

Answer 4

i don't know! i get three tries and that last answer was incorrect

Answer 5

amortized means payoff?
8% per year?

Answer 6

monthly payment is 244.13

Answer 7

how you calculate it?

Answer 8

Use the formula


\[M=\frac{Pr/12(1+r/12)^n}{(1+r/12)^n-1}\]

So I got (10000*(0.08/12)(1+0.08/12)^48)/((1+0.08/12)^48-1) = 244.13

Answer 9

thanx. never saw something like this. can you tell how this formula is called, so I can look on the wiki?

Answer 10

Derivation Of mortgage formula
P=initial balance
R= interest rate
W=monthly payment


\[1 \text{st} \text{Month}\text{ }= P(1+R) -W\]
\[2 \text{nd} \text{Month}\text{ }=((P(1+R) -W)(1+R))-W\]
\[P(1+R)^2-W(1+R)-W\]

\[3\text{rd} \text{Month}=\left(\left(P(1+R)^2-W(1+R)-W\right)(1+R)\right)-W\]
\[P(1+R)^3-W(1+R)^2-W(1+R)-W\]
\[N \text{th} \text{Month}= P(1+R)^N-W\left(1+(1+R)+(1+R)^2+\text{...}.(1+R)^{N-1}\right)\]
\[\text{Using} \space \text{Geometric} \space\text{Sum} \text{Formula}\]
\[\left(1+(1+R)+(1+R)^2+\text{...}.(1+R)^{N-1}\right)=\frac{\left(1-(1+R)^N\right)}{1-(1+R)}=-\frac{1-(1+R)^N}{R}\]
\[P(1+R)^N-W\left(-\frac{1-(1+R)^N}{R}\right)\]\[=P(1+R)^N+W\frac{\left(1-(1+R)^N\right) }{R}\]

Answer 11

W= monthly payment
R= Monthly payment
P= initial
\[10000\left(1+\frac{.08}{12}\right)^{48}+W\frac{\left(1-\left(1+\frac{.08}{12}\right)^{48}\right) }{\frac{.08}{12}}\text{==}0\]
W=244