In: Finance
A person deposits $60 at the end of each month in an account which earns 12.5% compounded monthly for 17 years. The person then stops making the deposits, but allows the money to remain in the bank earning the same interest for 5 more years. a. Find the value of this account at the end of 22 years. $ b. State the total amount of interest earned on this account. $
| a) | Step 1 :Value of annuity at the end of year 17 | |||||
| Future Value of an Ordinary Annuity | ||||||
| = C*[(1+i)^n-1]/i | ||||||
| Where, | ||||||
| C= Cash Flow per period | ||||||
| i = interest rate per period =12.5%/12 =1.041666667% | ||||||
| n=number of period =17*12 =204 | ||||||
| = $60[ (1+0.01041666667)^204 -1] /0.01041666667 | ||||||
| = $60[ (1.01041666667)^204 -1] /0.01041666667 | ||||||
| = $60[ (8.2814 -1] /0.01041666667] | ||||||
| = $41,940.70 | ||||||
| step 2 :Value of $41940.70 at the end of year 22 from now | ||||||
| FV= PV*(1+r)^n | ||||||
| Where, | ||||||
| FV= Future Value | ||||||
| PV = Present Value | ||||||
| r = Interest rate | ||||||
| n= periods in number | ||||||
| = $41940.7*( 1+0.01041667)^60 | ||||||
| =41940.7*1.86222 | ||||||
| = $78102.65 | ||||||
| b) | Total payment made = $60*12*17 | |||||
| =$12240 | ||||||
| Total interest earned = $78102.65-12240 | ||||||
| $ 65,862.65 | ||||||