John deposits dollar sign, 190$190 every month into an account earning an annual interest rate of 6% compounded monthly. How much would he have in the account after 11 years, to the nearest dollar? Use the following formula to determine your answer. A, equals, d, left bracket, start fraction, left bracket, 1, plus, i, right bracket, to the power n , minus, 1, divided by, i, end fraction, right bracket A=d( i (1+i) n −1 ) A, equalsA= the future value of the account after n periods d, equalsd= the amount invested at the end of each period i, equalsi= the interest rate per period n, equalsn= the number of periods
In this case:
d = $190
i = 6%/12 = 0.005 (monthly interest rate)
n = 11 years * 12 months/year = 132 months
Now we can plug these values into the formula:
A = $190 * ((0.005)*(1+0.005)^132 - 1) / 0.005
A = $190 * (0.005 * (1.005)^132 - 1) / 0.005
A = $190 * (0.005 * 2.3157 - 1) / 0.005
A = $190 * (0.0115785 - 1) / 0.005
A = $190 * (-0.9884215) / 0.005
A = $190 * -197.6843
A ≈ $37,360
Therefore, John would have approximately $37,360 in the account after 11 years.