# Nicole deposits \$350 in an account every month for 10 years. Determine the amount she will have at the end of 10 years when the interest rate is 3.7% per annum, compounded monthly.

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

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