The following is a reply to this thread from 2011.
This should be familiar (retrospective approach meaning PV if the reference point);
Time Payment Interest Principal Balance of Principal
1 Q Q(1-v^n) QV^n Qan|i
k Q Q(1-v(n-k+1)) QV^(n-k+1) Qa(n-k)|i
n Q Q(1-V) QV 0
12 monthly payments in 30 years gives n = 360
we are given k = 82 and QV^(360-82+1) = 259.34
when k = 56, QV^(360-56+1) = 230.19
Dividing these two we get V^-26 = 259.34/230.19
(1+i)^26 = 259.34/230.19; i = 0.004596489
use either k = 82 or k = 56 to solve for k we get;
QV^(360-82+1) = QV^279= 259.34; Q(1.004596489)^-279 = 259.34, Q = 932.27
last we are given K = 133 and told to calculate the amount of interest, so;
Q(1-V^(n-k+1)) = 932.27 (1-V^(360-133+1)) = 932.27(1-V^228) = 932.27(1-1.004596489^-228) = 604.59
Cheers!
				
			You'll find it's a good idea to "make up" a simple scenario, so you can "see" what's going on;
assume a $10,000 loan at 12% annual compounded monthly, over 9 months:
So we have:Code:p# payment principal interest balance 0 10000.00 1 1167.40 1067.40 100.00 8932.60 2 1167.40 1078.08 89.32 7854.52 3 1167.40 1088.86 78.54 6765.66 **** 4 1167.40 1099.75 67.65 5665.91 5 1167.40 1110.74 56.66 4555.17 6 1167.40 1121.85 45.55 3433.32 7 1167.40 1133.07 34.33 2300.25 8 1167.40 1144.40 23.00 1155.85 9 1167.40 1155.85 11.55 .00
A = amount borrowed ; 10000
n = number of payments ; 9
i = interest rate per month ; .12/12 = .01
P = monthly payment ; 1167.40
Let j = principal portion of payment
Let k = interest portion of payment
Now let's calculate the interest portion of the 3rd payment (t=3):
k = P(1 - v) where v = 1/(1 + i)^(n-t+1)
k = 1167.40(1 - v) where v = 1 / 1.01^7
k = 78.54 ****
Clear, right?
Now, if you'll repost your problem CLEARLY, and using my variables (hate yours!),
plus showing HOW you calculated whatever you got, I'll be happy to help.
This should be familiar (retrospective approach meaning PV if the reference point);
Time Payment Interest Principal Balance of Principal
1 Q Q(1-v^n) QV^n Qan|i
k Q Q(1-v(n-k+1)) QV^(n-k+1) Qa(n-k)|i
n Q Q(1-V) QV 0
12 monthly payments in 30 years gives n = 360
we are given k = 82 and QV^(360-82+1) = 259.34
when k = 56, QV^(360-56+1) = 230.19
Dividing these two we get V^-26 = 259.34/230.19
(1+i)^26 = 259.34/230.19; i = 0.004596489
use either k = 82 or k = 56 to solve for k we get;
QV^(360-82+1) = QV^279= 259.34; Q(1.004596489)^-279 = 259.34, Q = 932.27
last we are given K = 133 and told to calculate the amount of interest, so;
Q(1-V^(n-k+1)) = 932.27 (1-V^(360-133+1)) = 932.27(1-V^228) = 932.27(1-1.004596489^-228) = 604.59
Cheers!
			
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