Question

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)(1.0025n ? 1).(Round your answers to two decimal places.)

(a) Compute the first six terms of the sequence {An}.
A1 = $
A2 = $
A3 = $
A4 = $
A5 = $
A6 = $
(b) Find the balance in the account after 5 years by computing the 60th term of the sequence.
A60 =
(c) Find the balance in the account after 15 years by computing the 180th term of the sequence.
A180 =

Answer 1

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