You borrowed $5,000 from your parents to purchase a used car. You have agreed to make payments of $250 per month plus an additional 1% interest on the unpaid balance of the loan.
a. Is this problem an example of a geometric series or an arithmetic series? Support your answer mathematically by applying the concepts from this unit.
b. Find the first year’s monthly payments that you will make and the unpaid balance after each month.
c. Find the total amount of interest paid over the term of the loan.

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- amistre64

Does that mean that the 1% is compounded monthly?

- anonymous

That's kinda what it sounds like to me....

- amistre64

first payment of 250 + 5000(.01) = 300; remaining balance of 4750?

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## More answers

- anonymous

So 300 a month? I can figure the different left over balances after each month I think. How would I find the total amoun of interest paid just add up the percentages until I reach zero for my payments?

- amistre64

thats the long way to do it :) set up payment coupons of 250 + unpaid(.1)...but I think this is just a loan that is compounded monthly, just not too sure :)

- amistre64

5000/250 would be the number of times needed to pay the principal off

- anonymous

oh ok so it would take 20 payments?

- amistre64

at 20 payments the principal is gone :)

- amistre64

5000(.01) + (5000-250)(.01) + (5000 - 500)(.01).... for 20 times

- amistre64

(5000 - n(250))(.01)? does that sound right?

- anonymous

Ok so this is sorta making sense.... when you say for 20 times do I multiply all of that by 20?

- anonymous

n being 20 right?

- amistre64

lol.... I wish I knew fer sure :)
We know it takes 20 payments to pay that loan off; and each payment made subtracts 250 from the principal..

- anonymous

right

- amistre64

so there is a relation here that happens 20 times....

- anonymous

ok I understand that

- amistre64

im glad one of us does :)

- anonymous

lol well I think I do anyways sorta ha ha

- amistre64

first iteration is: (5000 - 0(250))(.01) = interest paid
second iteration: (5000 - 1(250))(.01) = interest paid
third iteration: (5000 - 2(250))(.01) = interset paid
.....

- anonymous

oh ok I was trying to but 20 in for n and was getting 0 which I knew wasn't right

- amistre64

lets clean this up: x = 5000 ; m=250 ; t = .01
xt - mt(n-1) = interest paid in any given iteration...how do we add these all up?

- anonymous

so the first interest paid is 50
second interest paid is 47.50
third interest paid is 45

- anonymous

hold on here we are trying to find the answer to question c right

- amistre64

good, good.....
theres an answer to find? :)

- amistre64

yes..c = total interest paid

- amistre64

if we can determine how much interest is paid for any given number of months, we will know the answer to "b"

- anonymous

ohhh ok so I have to find the interest paid for 12 months

- amistre64

yes.... and there is an equation you can use, just trying to figure it out :)

- amistre64

(xt - mt(0)) + (xt - mt(1)) + (xt - mt(2)) + (xt - mt(3))...
xt - mt(0) + xt - mt(1) + xt - mt(2) + xt - mt(3)....
what can we factor out of these?

- amistre64

xt (-mt(0)- mt(1) - mt(2) - mt(3) -mt(4))...

- anonymous

oh gosh I'm terrible at factoring

- amistre64

xt (-mt (n-1)) looks to be it; for any given number of months...
x = 5000 ; m=250 ; t = .01
50 (-2.5 (n-1)) sounds right

- amistre64

not quite there.... at n=1 we get 0 for interest....

- amistre64

whats the total interest we know for the first 3?

- amistre64

142.50 is the first 3 months of interest:
50 (-2.5 (3-1)) = 250 .... so it aint right yet :)

- amistre64

xt - mt(0) + xt - mt(1) + xt - mt(2) + xt - mt(3)....
xt(1 - mt(n-1) + 1 mt(n-1) + 1 - mt(n-1))...

- amistre64

100 (1 - 250(n-1)) when n=3
nope.... i guess we need someone smarter than me :)

- amistre64

2.5(21-n) will give you the amount of interest paid on a certain iteration..... so far, now just need to integrate that with respect to n ...maybe :)

- amistre64

number of iterations:

Amount of interest this payment: Thatll do it :)

Amount of interest this payment: Thatll do it :)

- amistre64

c) is $525 of interest is paid at the end of the loan (20 months).
b):
jan: 250+ 50 = 300 ; 4750
feb: 250+47.50 = 297.50 ; 4500
mar: 250+45 = 295 ; 4250
apr: 250+42.50 = 292.50 ; 4000
may: 250+40 = 290 ; 3750
jun: 250+37.50 = 287.50 ; 3500
jul: 250+35 = 285 ; 3250
aug: 250+32.50 = 282.50 ;3000
sep :250+30 = 280 ; 2750
nov: 250+27.50 = 277.50 ;2500
dec: 250+25 =275 ; 2250

- amistre64

a) id guess arithmetical series....

- anonymous

missing october in part b.... If it's arithmetic though I think I just do
Oct:250+27.50 = 277.50 ;2500
Nov:250+25 =275 ; 2250
Dec:250+22.50=272.5; 2000

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