Question

What would a monthly payment be on a purchase of a $11,000 car at 6.9% for 3 years?
A) $334.39
B) $339.15
C) $344.19
D) -$344.19
@thesmartone

Answer 1

apply this formula \[A = p (1 + \frac{ r }{ n })^{n \times t}\]
key word " monthly payment "
r =6.9% change it to decimal
t = time /years
p= initial amount

Answer 2

not quite the correct formula, at least not for a straight forward solution

Answer 3

if we wanted to know the amount of a current value; that gets compounded over a time span at a stated interest rate... then your formula is valid

Answer 4

otherwise we want to know when the geometric sum of payments balances with the loan at its compounding interest

\[Bk^n=P\frac{k^n-1}{k-1}\]
\[P=Bk^n\frac{k-1}{k^n-1}\]
which then simplifies to the usual textbook conflagrated formula for payment

Answer 5

11000(1+.069/12)^(36)(.069/12)/((1+.069/12)^(36)-1)

http://www.wolframalpha.com/input/?i=11000%281%2B.069%2F12%29%5E%2836%29%28.069%2F12%29%2F%28%281%2B.069%2F12%29%5E%2836%29-1%29

Answer 6

@amistre64 may you please do the problem? I dont know how to do those formulas

Answer 7

you just plug in the information is all .... i dont use the textbook formulas, i make my own

Answer 8

if you wanted to know how much youd have to pay at the end of 3 years making NO payments ... then
\[A=P(1+r/12)^{nt}\]
is good, but thats not what the question asks for

Answer 9

@amistre64 please?

Answer 10

i don't understand the 2nd part http://prntscr.com/6t1rbb
and never used that before
btw thanks @.amistre64

Answer 11

@nnesha can you give me the answer please?

Answer 12

the answer is already given ... its right there in plain site

now for the second part of it, lets assume we have a balance that needs to be paid off

Answer 13

the amount of the balance at any given time frame is: Bn

B0 = B ; the staring balance is the starting balance

B1 = Bk ; it compounds at some periodic rate

B1 = Bk - P ; then we subtract a payment

this process continues at regular intervals:

B2 = B1k - P, but we can determine B1 already

B2 = (Bk-P)k -P ; distribute and we get

B2 = Bk^2 -Pk -P ; a few terms and we will notice that the original balance is compounding, and each subsequent payment is affected as well, it part of the old balance

we get the setup

Bn = Bk^n - P - Pk - Pk^2 - Pk^3 - ... - Pk^(n-1)


factor out the -P and we have a geometric series in k

Bn = Bk^n - P( 1+k+k^2+k^3+...+k^(n-1)) ; which has a well know form

Answer 14

\[B_n=Bk^n-P\frac{k^n-1}{k-1}\]
this works for a variety of financial applications, annuity, loans, payments, ballons, etc ...

to find P, well we stop payments when Bn = 0
\[0=Bk^n-P\frac{k^n-1}{k-1}\]
\[Bk^n=P\frac{k^n-1}{k-1}\]
so this means that the geometric sum of payments has finally balanced out the compounding part of the loan.

Answer 15

ohhh okay thank you so much! :-)

Answer 16

youre welcome