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anonymous

  • 5 years ago

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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  1. amistre64
    • 5 years ago
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    Does that mean that the 1% is compounded monthly?

  2. anonymous
    • 5 years ago
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    That's kinda what it sounds like to me....

  3. amistre64
    • 5 years ago
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    first payment of 250 + 5000(.01) = 300; remaining balance of 4750?

  4. anonymous
    • 5 years ago
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    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?

  5. amistre64
    • 5 years ago
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    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 :)

  6. amistre64
    • 5 years ago
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    5000/250 would be the number of times needed to pay the principal off

  7. anonymous
    • 5 years ago
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    oh ok so it would take 20 payments?

  8. amistre64
    • 5 years ago
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    at 20 payments the principal is gone :)

  9. amistre64
    • 5 years ago
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    5000(.01) + (5000-250)(.01) + (5000 - 500)(.01).... for 20 times

  10. amistre64
    • 5 years ago
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    (5000 - n(250))(.01)? does that sound right?

  11. anonymous
    • 5 years ago
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    Ok so this is sorta making sense.... when you say for 20 times do I multiply all of that by 20?

  12. anonymous
    • 5 years ago
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    n being 20 right?

  13. amistre64
    • 5 years ago
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    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..

  14. anonymous
    • 5 years ago
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    right

  15. amistre64
    • 5 years ago
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    so there is a relation here that happens 20 times....

  16. anonymous
    • 5 years ago
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    ok I understand that

  17. amistre64
    • 5 years ago
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    im glad one of us does :)

  18. anonymous
    • 5 years ago
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    lol well I think I do anyways sorta ha ha

  19. amistre64
    • 5 years ago
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    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 .....

  20. anonymous
    • 5 years ago
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    oh ok I was trying to but 20 in for n and was getting 0 which I knew wasn't right

  21. amistre64
    • 5 years ago
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    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?

  22. anonymous
    • 5 years ago
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    so the first interest paid is 50 second interest paid is 47.50 third interest paid is 45

  23. anonymous
    • 5 years ago
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    hold on here we are trying to find the answer to question c right

  24. amistre64
    • 5 years ago
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    good, good..... theres an answer to find? :)

  25. amistre64
    • 5 years ago
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    yes..c = total interest paid

  26. amistre64
    • 5 years ago
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    if we can determine how much interest is paid for any given number of months, we will know the answer to "b"

  27. anonymous
    • 5 years ago
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    ohhh ok so I have to find the interest paid for 12 months

  28. amistre64
    • 5 years ago
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    yes.... and there is an equation you can use, just trying to figure it out :)

  29. amistre64
    • 5 years ago
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    (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?

  30. amistre64
    • 5 years ago
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    xt (-mt(0)- mt(1) - mt(2) - mt(3) -mt(4))...

  31. anonymous
    • 5 years ago
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    oh gosh I'm terrible at factoring

  32. amistre64
    • 5 years ago
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    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

  33. amistre64
    • 5 years ago
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    not quite there.... at n=1 we get 0 for interest....

  34. amistre64
    • 5 years ago
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    whats the total interest we know for the first 3?

  35. amistre64
    • 5 years ago
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    142.50 is the first 3 months of interest: 50 (-2.5 (3-1)) = 250 .... so it aint right yet :)

  36. amistre64
    • 5 years ago
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    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))...

  37. amistre64
    • 5 years ago
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    100 (1 - 250(n-1)) when n=3 nope.... i guess we need someone smarter than me :)

  38. amistre64
    • 5 years ago
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    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 :)

  39. amistre64
    • 5 years ago
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    <html> <body> number of iterations: <input id="inp1"><br><br> <input type = button value ="Calculate Total interest" onClick="qwe()"><br><br> <input id="inp2" value="total"><br><br> Amount of interest this payment:<input id="inp3"> </body> <script language=javascript> var j, i=0, t=0, x,c function qwe() { x = document.getElementById("inp1").value x=x*1 for (j=1;j<x+1;j=j+1) { i=2.5*(21-j) t=t+i } document.getElementById("inp2").value=t document.getElementById("inp3").value=i t=0 } </script> </html> Thatll do it :)

  40. amistre64
    • 5 years ago
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    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

  41. amistre64
    • 5 years ago
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    a) id guess arithmetical series....

  42. anonymous
    • 5 years ago
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    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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