annuities

[dcgirl]

i have a question where a person wants to save up $10 000 for a down payment by depositing $450 at the end of each month in an account paying 8%/a compounded monthly. This is how i have the equation sent up
10 000=450(1.066)^-n(1.066^n-1)/0.066
Is this right and if it is how do i solve for n?

 

[Unco]

The formula for the balance, B(n), at the end of the nth month (with zero principal) would be

\(\displaystyle \L\mbox{B(n) = \frac{c}{r}\left(\left(1 + r\right)^{n+1} - (1 - r)\right)}\)

where c is the amount being added at the end of each month, and r is 0.667 (or 1/150) as you have it.

I'm seeing some similarities and some differences between this and your equation.

Once you have the correct equation, solve by simplifying and isolating the (151/150)^(n+1) term, and taking logs of both sides.


As a side, you could also use the sum of a geometric series, with first term 450*1.0667.

 

[dcgirl]

im trying to do it as the sum of a geometric seris is this what it woul look like if i did that 10 000=450(1.066)^-n(1.066^n-1)/0.066

 

[Unco]

I still can't tell what you're doing.

The sum of the first n terms of a geometric series is given by

\(\displaystyle \L\mbox{ S_n = \frac{a\left(r^n - 1\right)}{r - 1}}\)

where a is the first term, which I gave previously.

Plug-n-chug.

 

[Denis]

i have a question where a person wants to save up $10 000 for a down payment by depositing $450 at the end of each month in an account paying 8%/a compounded monthly. This is how i have the equation sent up
10 000=450(1.066)^-n(1.066^n-1)/0.066
Is this right and if it is how do i solve for n?

First, it's 1.0066, not 1.066.

Equation: f = d[(1 + i)^n - 1] / i
f = future value (10000)
d = monthly deposit (450)
i = monthly interest rate (.08 / 12 = .0066...)
n = number of months (?)

10000 = 450(1.0066^n - 1) / .0066
1.0066^n = 10000(.0066)/450 + 1
n = log[10000(.0066)/450 + 1] / log(1.066) ...you should get close to 21 months