Question

Meghan has a 15-year adjustable rate mortgage with a rate of 5.4% for the first 5 years. The monthly payment is $2,394.77. The amount of the mortgage is $295,000. What is the remaining balance after 5 years rounded to the nearest dollar?

$253,392

$151,314

$221,673

$196,667
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I GOT $431,058.69 WHAT SHOULD DO FROM THERE ?

Answer 1

Do I use this FVn = P[(1+c)n - 1]/c

Answer 2

Balance time the (compounding set)^5; minus the payments times the geometric sum of the compunding setup

Answer 3

i tend to be forgetting the scalar of 12 for some reason
\[B_5=B_ok^{5*12}-P~\frac{1-k^{5*12}}{1-k}~:~k=1+\frac{.054}{12}\]

Answer 4

http://www.wolframalpha.com/input/?i=295000%281%2B.054%2F12%29%5E%2812*5%29-%282394.77%29%281-%281%2B.054%2F12%29%5E%2812*5%29%29%2F%281-%281%2B.054%2F12%29%29

Answer 5

So use the formula you used if I run into this problem again ?

Answer 6

i tend to use that formula for all of it; i have to modify it at times tho

Answer 7

whats the formula called ?

Answer 8

dunno, i made it for myself ...

Answer 9

lol can I get the original form of it to remember ?

Answer 10

if we run thru what happens when you make payments on a balance:\[B_1=B_o(1+\frac r{12})-P\]
\[B_2=B_1(1+\frac r{12})-P\]but replacing the value of B1 we get\[B_2=(B_0(1+\frac r{12})-P)(1+\frac r{12})-P\]
for clean up lets say that 1+r/12 is k and continue

\[B_2=(B_0(k)-P)k-P=B_0(k)^2-P(1+k)\]
\[B_3=B_0(k)^3-P(1+k+k^2)\]
\[B_4=B_0(k)^4-P(1+k+k^2+k^3)\]
\[B_n=B_0(k)^n-P(1+k+k^2+k^3+...+k^{n-1})\]

Answer 11

that tag along to the payments is a geometric sum with a formula of\[\frac{1-k^n}{1-k}\]
therefore\[B_n=B_o(k)^n-P\frac{1-k^n}{1-k}\]

Answer 12

this is the original form

Answer 13

the n represent the number of periods, which tends to be adjusted as 12*number of years

Answer 14

using this form i can calculate the payments need to zero out the balance after so many years by putting Bn = 0 and solving for P

Answer 15

i can calculate annuities with it as well

Answer 16

i can even use it to restructure thos ARM adjustable rate contrivances

Answer 17

thank you soo much

Answer 18

youre welcome, ive seen what the textbooks offer as formulas and they simply are not intuitive to me

Answer 19

im confused @amistre64

Answer 20

i use my own formula for this .... its one a created. But it works :)

Answer 21

so wait... i got 221674, but thats not an optioon

Answer 22

?

Answer 23

221674 is an approximation ... 221673 is the closest one to it in the options

Answer 24

oh, so its that one?

Answer 25

yes

Answer 26

the actual amount is: 221673.51

Answer 27

kk ahhhhhhhhhhhhhhh