3. A periodic deposit is made into an annuity with the given terms. Find how much the annuity will hold at the end of the specified amount of time. Round your answer to the nearest dollar.Regular deposit:$1900Interest rate:2.4%FrequencyquarterlyTime:17 yearsFuture value: $

3 A periodic deposit is made into an annuity with the given terms Find how much the annuity will hold at the end of the specified amount of time Round your answ class=

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The formula for the future value of annuity is given by

Where,

[tex]\begin{gathered} A=\text{regular deposits= \$1900} \\ r=2.4\text{ \%=}\frac{\text{2.4}}{100}=0.024 \\ T=\text{ number of years =17} \\ m=\text{frequency}=4 \end{gathered}[/tex]

By substituting the values, we will have

[tex]FV_{OA}=A\lbrack(\frac{1+\frac{r}{m})^{mT}-1}{\frac{r}{m}}\rbrack[/tex]

Therefore,

We will have

[tex]\begin{gathered} FV_{OA}=A\lbrack(\frac{(1+\frac{r}{m})^{mT}-1}{\frac{r}{m}}\rbrack \\ FV_{OA}=1900\frac{\lbrack(1+\frac{0.024}{4})^{4\times17})_{}-1\rbrack}{\frac{0.024}{4}} \end{gathered}[/tex][tex]\begin{gathered} FV_{OA}=1900\frac{\lbrack(1+\frac{0.024}{4})^{4\times17})_{}-1\rbrack}{\frac{0.024}{4}} \\ FV_{OA}=\frac{1900\lbrack(1+0.006)^{68}-1\rbrack}{0.006} \\ FV_{OA}=\frac{1900\lbrack(1.006)^{68}-1\rbrack}{0.006} \\ FV_{OA}=\frac{1900(0.501975)}{0.006} \\ FV_{OA}=\frac{953.7525}{0.006} \\ FV_{OA}=\text{ \$}158,958.75 \\ \text{Appro}\xi\text{mately to the nearest dollar will be} \\ FV_{OA}=\text{ \$158,959} \end{gathered}[/tex]

Therefore,

The future value of the annuity = $158,959

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