# An investor deposists $10,000 in an account that earns 3.5% interest compounded quarterly. The balance in the account after n quarters is given by `A_n = 10,000(1 + 0.035/4)^n , n = 1, 2, 3 ...`...

An investor deposists $10,000 in an account that earns 3.5% interest compounded quarterly. The balance in the account after n quarters is given by

`A_n = 10,000(1 + 0.035/4)^n , n = 1, 2, 3 ...`

a) Write the first terms of the sequence

b) Find the balance in the account after 10 years by computing the 40th term of the sequence.

c) Is the balance after 20 years twice the balance after 10 years? explain.

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### 1 Answer

a.) For the first terms of the sequence we will assign numbers to n.

Substitute `1` for `n ` for the first term.

`A_1 = 10,000(1 + (0.035)/(4))^1`

Do what is in the parentheses first.

`A_1 = 10,000(1 + .00875)^1 = 10,000(1.00875)^1`

Then, consider the doing the exponent part.

`A_1 = 10,000(1.00875)^1 = 10,000(1.00875)`

Do the multiplication.

`A = 10,000(1.00875) = 10,080`

Substitute `2` for `n` for the second term.

`A_2 = 10,000(1 + (0.035)/(4))^2`

Do what is in the parentheses first.

`A_2 = 10,000(1 + .00875)^2 = 10,000(1.00875)^2`

Then, consider the doing the exponent part.

`A_2 = 10,000(1.00875)^2 = 10,000(1.017576563)`

Do the multiplication.

`A_2 = 10,000(1.017576563) = 10,175.77.`

Substitute `3` for `n` for the third term.

`A_3 = 10,000(1 + (0.035)/(4))^3`

Do what is in the parentheses first.

`A_3 = 10,000(1 + .00875)^3 = 10,000(1.00875)^3`

Then, consider the doing the exponent part.

`A_3 = 10,000(1.00875)^3 = 10,000(1.026480357)`

Do the multiplication.

`A = 10,000(1.026480357) = 10,264.80.`

Therefore, the first `3` terms are `$10,080` `, $10,175.77` and `$10,264.80` .

b.) We plug-in `n = 40` .

`A_40 = 10,000(1 + (0.035)/(4))^40`

`= 10,000(1 + .00875)^40 `

`= 10,000(1.00875)^40 `

`= 10,000(1.416908838)`

`A_40 = 14,169.09.`

Therefore, the balance in the account after `10` years is `$14,169.09` .

c.) The balance after` 20` years will not be twice the balance after `10` years,

because the interest rate is just `3.5%` and it is compounded quarterly.

We can try finding the account balance after `20` years, we will plug-in `80` to the` n` ,

since `20 * 4 = 80` , we multiply by` 4` since it is compounded quarterly.

`A_80 = 10,000(1 + (0.035)/(4))^80`

`= 10,000(1 + .00875)^80 `

`= 10,000(1.00875)^80 `

`= 10,000(2.007630655)`

`A_80 = 20076.31`

The account balance after `10` years is `$14,169.09` . If that is doubled it should be,

`28338.18` . However, we only got `$20,076.31` for the account balance after `20` years.