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Consider making monthly deposits of P= \$55 in a savings account at an annual interest rate r=2.5...

Question:

Consider making monthly deposits of {eq}P= \$55{/eq} in a savings account at an annual interest rate {eq}r=2.5 \%{/eq}. Find the balance {eq}A{/eq} after {eq}t=30{/eq} years when the interest is compounded as follows.

(a) monthly

{eq}A = P(\frac{12}{r})[(1 + \frac{r}{12})^{12t} -1]{/eq}

(continously)

{eq}A = \frac{P(e^{rt} -1)}{e^{r/12} -1}{/eq}

Annuity:

The following formula is used for calculating the future value when there are monthly deposits and interest is credited compounded monthly:

{eq}A = P(\dfrac{12}{r}) \left [ (1 + \dfrac{r}{12})^{12t} -1 \right ] {/eq}

Where, P is the amount of deposit, r is the annual rate of interest and t is the time in years.

The following formula is used for calculating the future value when there are monthly deposits and interest is credited compounded continuously:

{eq}A = \dfrac{P(e^{rt} -1)}{e^{r/12} -1} {/eq}

Where, P is the amount of deposit, r is the annual rate of interest and t is the time in years.

Answer and Explanation:

Given,

Monthly deposits:

{eq}P = \$55 {/eq}

Annual rate of interest:

{eq}r = 2.5\% = 0.025 {/eq}

Time in years:

{eq}t = 30 {/eq}

(a)

Finding the balance when interest is compounded monthly by using the formula:

{eq}A = P(\dfrac{12}{r}) \left [ (1 + \dfrac{r}{12})^{12t} -1 \right ] \\ A = 55(\dfrac{12}{0.025}) \left [ (1 + \dfrac{0.025}{12})^{12(30)} -1 \right ] \\ A = 26400 \left [ (\dfrac{12.025}{12})^{360} -1 \right ] \\ \boxed{A \approx 29445.21}. {/eq}

(b)

Finding the balance when interest is compounded continuously by using the formula:

{eq}A = \dfrac{P(e^{rt} -1)}{e^{r/12} -1} \\ A = \dfrac{55(e^{0.025(30)} -1)}{e^{0.025/12} -1} \\ A = \dfrac{55(e^{0.75} -1)}{e^{0.025/12} -1} \\ \boxed{A \approx 29458.09}. {/eq}


Learn more about this topic:

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How to Find the Value of an Annuity

from Algebra II Textbook

Chapter 21 / Lesson 15
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