find the value of a retirement savings account paying an apr of 6.1% (compounded monthly) after 25 years of…

Explanation:

Step1: Identify the formula for future - value of an ordinary annuity

The formula for the future - value of an ordinary annuity is $F = P\times\frac{(1 + r)^{n}-1}{r}$, where $P$ is the monthly payment, $r$ is the monthly interest rate, and $n$ is the total number of payments. The annual percentage rate (APR) is $6.1%=0.061$. The monthly interest rate $r=\frac{0.061}{12}$. The number of years is 25, so the number of payments $n = 25\times12=300$.

Step2: Calculate for $P = 61.50$

Substitute $P = 61.50$, $r=\frac{0.061}{12}$, and $n = 300$ into the formula. First, calculate $(1 + r)^{n}=(1+\frac{0.061}{12})^{300}$. Let $x=\frac{0.061}{12}\approx0.0050833$. Then $(1 + x)^{300}=(1 + 0.0050833)^{300}$. Using the formula $a^{b}=e^{b\ln(a)}$, we have $(1 + 0.0050833)^{300}=e^{300\ln(1.0050833)}$. $\ln(1.0050833)\approx0.00507$, and $300\times0.00507 = 1.521$. So $(1 + 0.0050833)^{300}=e^{1.521}\approx4.577$. $(1 + r)^{n}-1\approx4.577-1 = 3.577$. $\frac{(1 + r)^{n}-1}{r}=\frac{3.577}{\frac{0.061}{12}}=\frac{3.577\times12}{0.061}\approx698.295$. $F=P\times\frac{(1 + r)^{n}-1}{r}=61.50\times698.295\approx42845.14$.

Answer:

$42845.14$