# Amortization Formula :( Pleez?

• Nov 9th 2008, 07:28 PM
Steph07
Amortization Formula :( Pleez?
Thank you fo helping me, if you can lol (Worried):

Use the Amortization Formula to compute
a monthly car payment if L = $26,000 financed over 60 months at 4.9% interest rate. Payment = Loan (x 1 - 1 (1-(1+x)^N) Payment = L(x(1-1÷(1-(1+x)^N))) Code for Symbols r≡annual interest in decimal format y≡number of payments to be made per year (integer) N≡total number of payments to be made (integer) L≡amount of Loan x≡r÷y ()≡(parentheses) ^≡exponent caret • Nov 10th 2008, 10:42 AM masters Quote: Originally Posted by Steph07 Thank you fo helping me, if you can lol (Worried): Use the Amortization Formula to compute a monthly car payment if L =$26,000 financed
over 60 months at 4.9% interest rate.

Payment = Loan (x 1 - 1
(1-(1+x)^N)

Payment = L(x(1-1÷(1-(1+x)^N)))
Code for Symbols
r≡annual interest in decimal format
y≡number of payments to be made per year (integer)
N≡total number of payments to be made (integer)
L≡amount of Loan
x≡r÷y
()≡(parentheses)
^≡exponent caret

I can't really tell what your formula says.
I used the following amortization formula found Here.

$A=P\frac{r(1+r)^n}{(1+r)^n-1}$, where

where

• A = payment Amount per period (monthly periods)
• P = initial Principal (loan amount)
• r = interest rate per period (yearly rate / 12)
• n = total number of payments or periods
A = payment per month
P = 26,000
r = 4.9% per year / 12 months = $\frac{.049}{12}$%
n = 60 monthly periods

$A=26000\frac{\frac{.049}{12}\left(1+\frac{.049}{12 }\right)^{60}}{\left(1+\frac{.049}{12}\right)^{60}-1}$

$A=\489.46$