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January 23rd, 2013, 10:56 AM  #11  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: APR with fee payment Quote:
Code: 0 1500.00 1 510.03 15.00 1004.97 2 510.03 10.05 504.99 3 510.03 5.04 .00 required monthly payment = $510.03, 1st payment 1 month after loan obtained. Yours: pR(R^n  1) / (R  1) = 510.03(1.01)(1.01^3  1) / (1.01  1) = ~1561 Mine: p(1  1/R^n) / (R  1) = 510.03(1  1/1.01^3) / (1.01  1) = ~1500 So me is confused...  
January 23rd, 2013, 01:07 PM  #12 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: APR with fee payment
To JMATH17; to illustrate reference to geometric series... Take the $1,500 loan from my previous post. It all works this way: assume no payments on loan, instead the payments are EACH put in a savings account at same rate as loan: Code: n int balance int balance int balance int balance 0 .00 1500.00 1 15.00 1515.00 .00 510.03 2 15.15 1530.15 5.10 515.13 .00 510.03 3 15.30 1545.45 5.16 520.29 5.10 515.13 .00 510.03 1500(1 + .01)^3 = 510.03(1 + .01)^2 + 510.03(1 + .01)^1 + 510.03(1 + .01)^0 Let R = 1 + .01 ; rewrite above: AR^n = pR^(n1) + pR^(n2) + pR^(n3) Rearrange, and perform following: Code: AR^n = pR^0 + pR^1 + pR^2 + ...... + pR^(n1) AR^n * R = pR^1 + pR^2 + ...... + pR^(n1) + pR^n ================================================================ AR^n * R  AR^n = pR^0 + 0 + 0 + ...... + 0 + pR^n A = p(R^n  1) / [R^n(R  1)] Substituting back R = 1+i and simplifying: A = [1  1/(1+i)^n] / i Hope that helps... 
January 23rd, 2013, 05:18 PM  #13  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: APR with fee payment Quote:
Where this you get this problem anyway? 2 years later, owing will be 5000(1 + .065/12)^24 = 5692.14 So effectively a loan of that amount will be obtained then, at 6.5 rate...  

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