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Seth has a bank account which pays 1.01% interest, compounded quarterly. Seth withdraws $4,567 from the account every quarter for 35 years. Assuming that Seth does not make any deposits into this account and that the withdrawals occur at the end of every quarter, find the initial value of the account, rounded to the nearest cent.




a.

$765,824.68


b.

$767,758.39


c.

$538,021.66


d.

$539,380.16






Please select the best answer from the choices provided




A

B

C

D

Respuesta :

nobillionaireNobley
[tex]A=P \frac{1-(1+ \frac{r}{t} )^{-nt}}{ \frac{r}{t} } [/tex]; where A is the initial value, P is the periodic withdrawal, r is the rate, t is the number of compounding in a year, n is the number of years.

[tex]A=P \frac{1-(1+ \frac{r}{t} )^{-nt}}{ \frac{r}{t} } \\ =4,567 \times \frac{1-(1+ \frac{0.0101}{4} )^{-35 \times 4}}{ \frac{0.0101}{4} } \\ =4,567 \times \frac{1-(1+ 0.002525)^{-140}}{ 0.002525 } \\ =4,567 \times \frac{1-(1.002525)^{-140}}{ 0.002525 } \\ =4,567 \times \frac{1-0.7025}{ 0.002525 } \\ =4,567 \times \frac{0.2975}{ 0.002525 } \\ =4,567 \times 117.8 \\ =$538,021.66[/tex]
Therefore, initial value = $538,021.66

Answer:

C. $538,021.66

Step-by-step explanation:

It is given that the money Seth withdraws was compounded every quarter for 35 years. So, we get,

Amount withdrawn every quarter, P = $4567

Rate of interest, r  = [tex]\frac{0.0101}{4}[/tex] = 0.002525

Time period, n = 35 × 4 = 140

Now, as we know the formula for annuity as,

[tex]P=\frac{r \times PV}{1-(1+r)^{-n}}[/tex]

where P = installments, PV = present value, r = rate of interest and n = time period.

This gives, [tex]PV=\frac{P \times [1-(1+r)^{-n}]}{r}[/tex]

i.e. [tex]PV=\frac{4567 \times [1-(1+0.002525)^{-140}]}{0.002525}[/tex]

i.e. [tex]PV=\frac{4567 \times [1-(1.002525)^{-140}]}{0.002525}[/tex]

i.e. [tex]PV=\frac{4567 \times [1-0.7021]}{0.002525}[/tex]

i.e. [tex]PV=\frac{4567 \times 0.2975}{0.002525}[/tex]

i.e. [tex]PV=\frac{1358.68}{0.002525}[/tex]

i.e. [tex]PV=538,091.08[/tex]

So, the closest answer to initial value of the account is $538,021.66

Hence, option C is correct.

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