Question

WILL MARK BRAINILEST!!!!!

Answer 1

To solve this we are going to use the compound interest formula with periodic deposits: $A=P(1+ \frac{r}{n} )^{nt}+P_{d}( \frac{(1+ \frac{r}{n})^{nt}-1 }{ \frac{r}{n} } )(1+ \frac{r}{n} )$
where 
$A$ is the final amount after $t$ years 
$P$ is the initial investment 
$P_{d}$ is the periodic deposits
$r$ is the interest rate in decimal form 
$n$ is the number of times the interest is compounded per year 
$t$ is the time in years

Since he is going to save from 27 years old  until 65 years old, $t=65-27=38$. We know that hes is opening his IRA with $0, so $P=0$; We also know that he is going to invest $200 at the beginning of each month, so $P_{d}=200$. To convert the interest rate to decimal form, we are going to divide it by 100: $r= \frac{2.65}{100} =0.0265$, and since the interest is compounded monthly, $n=12$. Lets replace all the values in our formula to find $A$:
$A=P(1+ \frac{r}{n} )^{nt}+P_{d}( \frac{(1+ \frac{r}{n})^{nt}-1 }{ \frac{r}{n} } )(1+ \frac{r}{n} )$
$A=0(1+ \frac{0.0265}{12} )^{(12)(38)}+200( \frac{(1+ \frac{0.0265}{12})^{(12)(38)}-1 }{ \frac{0.0265}{12} } )(1+ \frac{0.0265}{12} )$
$A=157419.04$

We can conclude that Rick will have $157,419.04 in his IRA account by the time when he retires.