Nicole deposits $350 in an account every month for 10 years. The amount she will have at the end of 10 years for an interest rate of 3.7% per annum, compounded monthly, has to be determined.

The rate of interest of the account is 3.7% and it is compounded monthly.

If an amount of $350 is deposited in the account, after n months it increases to `350*(1+0.037/12)^n = 350*(1+0.003083)^n = 350*1.003083^n`.

An amount of 350 invested at the beginning of 10 years is equal to `350*1.003083^120` after 10 years. The amount invested in the following month would increase to `350*1.003083^119` . The last $350 deposited increases to `350*1.003083^1` .

The sum of all the terms above gives the amount in her account at the end of 10 years.

`S = 350*1.003083^120 + 350*1.003083^119 + ... +350*1.003083^1`

= `350*(1.003083^120 + 1.003083^119 + ... + 1.003083^1)`

The sum of n terms of a geometric series `a, ar, ar^2... ar^(n-1)` is given by `a*(r^n - 1)/(r - 1)`

`350*(1.003083^120 + 1.003083^119 + ... + 1.003083^1)`

= `350*1.003083*(1.003083^(120-1))/(1.003083-1)`

= `350*1.003083*(1.003083^(120-1))/(1.003083-1)`

= 50,379.34.

At the end of 10 years, Nicole has $50,379.34 in her account.

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now