anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • amistre64
Does that mean that the 1% is compounded monthly?
anonymous
  • anonymous
That's kinda what it sounds like to me....
amistre64
  • amistre64
first payment of 250 + 5000(.01) = 300; remaining balance of 4750?

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anonymous
  • 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
  • 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
  • amistre64
5000/250 would be the number of times needed to pay the principal off
anonymous
  • anonymous
oh ok so it would take 20 payments?
amistre64
  • amistre64
at 20 payments the principal is gone :)
amistre64
  • amistre64
5000(.01) + (5000-250)(.01) + (5000 - 500)(.01).... for 20 times
amistre64
  • amistre64
(5000 - n(250))(.01)? does that sound right?
anonymous
  • anonymous
Ok so this is sorta making sense.... when you say for 20 times do I multiply all of that by 20?
anonymous
  • anonymous
n being 20 right?
amistre64
  • 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
  • anonymous
right
amistre64
  • amistre64
so there is a relation here that happens 20 times....
anonymous
  • anonymous
ok I understand that
amistre64
  • amistre64
im glad one of us does :)
anonymous
  • anonymous
lol well I think I do anyways sorta ha ha
amistre64
  • 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
  • anonymous
oh ok I was trying to but 20 in for n and was getting 0 which I knew wasn't right
amistre64
  • 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
  • anonymous
so the first interest paid is 50 second interest paid is 47.50 third interest paid is 45
anonymous
  • anonymous
hold on here we are trying to find the answer to question c right
amistre64
  • amistre64
good, good..... theres an answer to find? :)
amistre64
  • amistre64
yes..c = total interest paid
amistre64
  • amistre64
if we can determine how much interest is paid for any given number of months, we will know the answer to "b"
anonymous
  • anonymous
ohhh ok so I have to find the interest paid for 12 months
amistre64
  • amistre64
yes.... and there is an equation you can use, just trying to figure it out :)
amistre64
  • 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
  • amistre64
xt (-mt(0)- mt(1) - mt(2) - mt(3) -mt(4))...
anonymous
  • anonymous
oh gosh I'm terrible at factoring
amistre64
  • 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
  • amistre64
not quite there.... at n=1 we get 0 for interest....
amistre64
  • amistre64
whats the total interest we know for the first 3?
amistre64
  • amistre64
142.50 is the first 3 months of interest: 50 (-2.5 (3-1)) = 250 .... so it aint right yet :)
amistre64
  • 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
  • amistre64
100 (1 - 250(n-1)) when n=3 nope.... i guess we need someone smarter than me :)
amistre64
  • 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
  • amistre64
number of iterations:





Amount of interest this payment: Thatll do it :)
amistre64
  • 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
  • amistre64
a) id guess arithmetical series....
anonymous
  • 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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