Question

On January 1, 2013, $9500 is invested at a monthly rate of 1.5 % compounded monthly.
Write the sequence of four terms that gives the value of this investment on January 1 of each successive year.
January 1, 2014 $
January 1, 2015 $
January 1, 2016 $
January 1, 2017 $
After how many months is the investment first worth at least $39000?
(Remember that interest is only compounded at the end of the month.)
For the purposes of this problem, round all answers to the nearest cent.

Answer 1

Answer:

January 1, 2014 $9,643.48

January 1, 2015 $9,789.13

January 1, 2016 $9,936.99

January 1, 2017 $10,087.07

1131 months

Step-by-step explanation:

To calculate the value of the investment over the years, we can use the compound interest formula:

$\boxed{\begin{array}{l}\underline{\textsf{Compound Interest Formula}}\\\\A=P\left(1+\dfrac{r}{n}\right)^{nt}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\;\textsf{ A is the future value.}\\\phantom{ww}\bullet\;\;\textsf{ P is the principal investment.}\\\phantom{ww}\bullet\;\;\textsf{ r is the annual interest rate (in decimal form).}\\\phantom{ww}\bullet\;\;\textsf{ n is the number of times interest is applied per year.}\\\phantom{ww}\bullet\;\;\textsf{ t is the time (in years).}\end{array}}$

In this case:

• P = $9500
• r = 1.5% = 0.015
• n = 12

Substitute the values of P, r and n into the formula to create an equation for the future value in terms of t years.

$A=9500\left(1+\dfrac{0.015}{12}\right)^{12t}$

Simplify:

$A=9500\left(1.00125\right)^{12t}$

Now, calculate the value of the investment value on January 1 of each successive year:

January 1, 2014 (t = 1):

$A=9500\left(1.00125\right)^{12}=\ 9643.48$

January 1, 2015 (t = 2):

$A=9500\left(1.00125\right)^{2 \cdot 12}=\ 9789.13$

January 1, 2016 (t = 3)

$A=9500\left(1.00125\right)^{3 \cdot 12}=\ 9936.99$

January 1, 2017 (t = 4):

$A=9500\left(1.00125\right)^{4 \cdot 12}=\ 10087.07$

To find out after how many years the investment will be worth at least $39000, we can set up the inequality:

$9500\left(1.00125\right)^{12t} \geq 39000$

Solve for t:

$\begin{aligned}9500\left(1.00125\right)^{12t} &\geq 39000\\\\\left(1.00125\right)^{12t} &\geq \dfrac{39000}{9500}\\\\\left(1.00125\right)^{12t} &\geq \dfrac{78}{19}\\\\\ln\left(\left(1.00125\right)^{12t}\right) &\geq \ln\left(\dfrac{78}{19}\right)\\\\12t\ln\left(1.00125\right) &\geq \ln\left(\dfrac{78}{19}\right)\\\\t&\geq \dfrac{\ln\left(\dfrac{78}{19}\right)}{12\ln\left(1.00125\right)}\\\\t&\geq 94.21015549...\; \sf years\end{aligned}$

Convert this into months by multiplying the value of t by 12:

$t&\geq 1130.5218659...\;\sf months$

Therefore, the investment will first be worth at least $39000 after 1131 months.