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# How many years will it take to save 200000 if you place 1700 per month in an account that earns 7 percent compounded monthly?

Updated: 10/24/2022

Wiki User

15y ago

I get seven and a half years (90 months). If you have a spread sheet (or calculator) you can use the following formula for finding the amount accumulated "Pn" after "n" months at monthly compounded interest "i" and monthly savings "d" Pn = (d/i)[(1+i)^(n+1) - 1] the monthly interest rate is found from the annual rate "I" from ; i = (I+1)^(1/12) - 1. Using I = .07 (7%) , this gives i = .00565. In the Pn equation you can solve for any other variable you choose. However solving for "n" , which is really what you asked for, involves logarithims and with a spread sheet its easier to just try different "n's" and see what gets you \$200,000 for "Pn". Using this approach I got n = 90 months will get you about \$201,000. ------------------------------------------------------------ The previous answer is a good attempt at approaching the question intuitively. However, this question, though cryptically so, is a geometric series question. Consider the following: Let's say the person deposits 1700 at the beginning of the first month. After one month, the amount would be 1700*(1+0.07/12). At this point, the person deposits another 1700. One month after that, the first 1700 would gain two months' interest and become 1700*(1+0.07/12)2 and the 1700 deposited in the second month would accumulate one month's interest and become 1700*(1+0.07/12). The person deposits another 1700. Therefore, the following list represents the total amount of money the person will have: End of first month: 1700*(1+0.07/12) + 1700 End of second month: 1700*(1+0.07/12)2 + 1700*(1+0.07/12) + 1700 End of third month: 1700*(1+0.07/12)3 + 1700*(1+0.07/12)2 + 1700*(1+0.07/12) + 1700 ...... End of nth month: 1700*(1+0.07/12)n + 1700*(1+0.07/12)(n-1) + ... + 1700*(1+0.07/12)2 + 1700*(1+0.07/12) + 1700 Let's say that after nth month, the total amount is 200000 Now we recognize the formulat for the total amount of money at the end of the nth month is a geometric series with a = 1700 and r = (1+0.07/12) and n = n. Therefore: 200000 = 1700*(1+0.07/12)n + 1700*(1+0.07/12)(n-1) + ... + 1700*(1+0.07/12)2 + 1700*(1+0.07/12) + 1700 200000 = a(1-rn)/(1-r) 200000 = 1700(1-(1+0.07/12)n)/(1-(1+0.07/12)) n = log(1-200000*(-0.07/12)/1700)/log(1+0.07/12) n = 89.836 n ~ 90 months Additional Comments I would like to make two comments. First, regarding the second answer. Although .07/12 is approximately the monthly interest it does not give the correct value of 7% per annum because of the monthly compounding. Second, regarding the first answer. It is not an "intuitive" result. The equation for "Pn" is a rigorous derivation based on the same geometrical series analysis in the second solution. My intent was to spare the questioner the lengthy analysis required to arrive at the solution.

Wiki User

15y ago