April 10th, 2015, 05:08 PM  #31 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5  
April 10th, 2015, 09:06 PM  #32 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5 
I suppose the formula did provide the answer you were looking for The formula for sum of increasing payments (without interest) allows for any payment frequency per year My offer for free of cost help is still in place, so feel free to post more problems for which you seek a formula 
April 11th, 2015, 07:01 AM  #33 
Newbie Joined: Apr 2015 From: miami Posts: 18 Thanks: 0 
Yes it did indeed! the formulas you have helped me with will be very useful to me for my work. I can't thank you enough. 
April 11th, 2015, 08:06 AM  #34 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5 
No problem at all! Proud to have served the mentor, 22 degrees celsius of separation. Be at your service if work demands further problem solving 
April 11th, 2015, 08:52 AM  #35 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,955 Thanks: 988 
Abraham, aka Dexter, there was no need to change your original formula; (the one at post#24, showing FV of 13,481.60 ... and was CORRECT!) all that needed to be done was use i = 0; then 10,855.00 would result. However, the unnecessary extra work you did attempting to help your fellow man will be duly recorded in your Judgement Day file by Big Pete, seriously taken in consideration when your time in Purgatory is calculated, such calculation using simple interest, unfortunately i > 0. (there ain't no compounding up there, only winging or burning) 
April 11th, 2015, 12:47 PM  #36 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,955 Thanks: 988 
This is much easier if the "sum of arithmetic series" formula is used: S = n[2a + d(n  1)] / 2 a = first term d = difference between terms n = number of terms With your problem, I'll assume 2.75 years. a = 100 d = 15 n = 2 (number of complete years) f = 19 (number of deposits in final year) As far as f goes, that's easiest way to handle. 26 * .75 = 19.5 : so use 19 or 20. Ok to use 19.5, but doesn't make much sense to do so. If you want complete years only, simply use 0 for f. This handles complete years: 26n[2a + d(n1)] / 2 : result = 5590 (100*26 + 115*26) This handles final year: f(a + dn) : result = 2470 (19*130 = 2470) So: SUM = 26n[2a + d(n1)] / 2 + f(a + dn) : result = 5590 + 2470 = 8060. 

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