Question

In order to purchase a home, a family borrows $267,000 at 10.8% for 15 years. What is the monthly house payment necessary to amortize the loan? (Do not round until the final answer. Then, round to the nearest cent.)
a)$3,394.52
b)$2,403.00
c)$3,001.27
d)$19,108.39

Answer 1

Take a look at this: http://is.gd/mlAt6B

Answer 2

it looks like c

Answer 3

does monthly payments assume that the interest is compounded monthly?

Answer 4

yes

Answer 5

so, assuming monthly periods\[M_1=B(1+\frac i{12})-P\]
\[M_2=M_1(1+\frac i{12})-P\\~~~~~~=B(1+\frac i{12})^2-P(1+\frac{i}{12})-P\]
to clean things up, let k = 1+i/12
\[M_3=M_2(k)-P\\~~~~~~=B(k)^3-P(k)^2-P(k)-P\\~~~~~~=B(k)^3-P(k^2+k+1)\]
\[M_4=M_3(k)-P\\~~~~~~=B(k)^3-P(k^3+k^2+k+1)\]
noticing a pattern of the nth month and the exponents
\[M_n=M_{n-1}(k)-P\\~~~~~~=B(k)^n-P(k^{n-1}+...+k^3+k^2+k+1)\]
those ks attached to the Ps develop into a geometric sum
\[M_n=M_{n-1}(k)-P\\~~~~~~=B(k)^n-P(\frac{1-k^n}{1-k})\]
letting n = 18*12 months gives us 216 months
\[M_{216}=B(k)^{216}-P(\frac{1-k^{216}}{1-k})\]
letting M216 = 0
\[0=B(k)^{216}-P(\frac{1-k^{216}}{1-k})\]
\[B(k)^{216}=P(\frac{1-k^{216}}{1-k})\]
\[\frac{B(k)^{216}(1-k)}{1-k^{216}}=P\]

Answer 6

does that match up with the c answer? thats what i got with the calculator?

Answer 7

dunno yet

Answer 8

phone call, daughter issues :)

Answer 9

i get 2809, which is close enough to c to warrant variances in methods

Answer 10

what method did you use?

Answer 11

i used the calculator that someone provided

Answer 12

well, then as long as you hit the right buttons, ill assume its correct ....

Answer 13

http://www.bankrate.com/calculators/mortgages/amortization-calculator.aspx?ec_id=m1024777&ef_id=dCVP-HaRJnMAAACt%3a20130402193615%3as

yeah, i keep getting 2808.49 as a payment even when i use on online calculator

Answer 14

ok, thanks!

Answer 15

ill get this right sooner or later lol,

12 months times 15 years = 180 periods ....

so yeah, 3001 is fine