Complete questions:
A deposit of $200 is made in an account at the beginning of each month at an annual interest rate of 3% compounded monthly. The balance in the account after n months is An = 200(401)(1.0025^n - 1).(Round your answers to two decimal places.)
Answer:
Kindly check explanation
Explanation:
Given the amount of balance in an account after. 'n' months using the eqation:
An = 200(401)(1.0025^n - 1)
(a) Compute the first six terms of the sequence {An}.
A1 = 200(401)(0.0025) = $200.50
A2 = 200(401)(1.0025^2 - 1) = $401.50
A3 = 200(401)(1.0025^3 - 1) = $603.00
A4 = 200(401)(1.0025^4 - 1) = $805.01
A5 = 200(401)(1.0025^5 - 1) = $1007.53
A6 = 200(401)(1.0025^6 - 1) = $10210.54
B) balance 5 years after n = 5 * 12 = 60
A60 = 200(401)(1.0025^60 - 1) = $12,961.67
C.) balance 15 years after ; n = 15 * 12 = 180
A180 = 200(401)(1.0025^180 - 1) = $45,508.02