If Jerry contributes at the beginning of the month and withdraws at the end of the month, the final contribution earns 1 month's interest. The one before that earns 2 months' interest, so has a value of (1+0.017/12) times that of the last payment. In short, the sum is that of a geometric sequence with first term
a₁ = 300*(1+0.017/12)
and common ratio
r = 1+0.017/12
We assume Jerry contributes each month for 15 years, so a total of 180 payments. The sum is given by the formula for the sum of a geometric sequence.
![S_{n}=a_{1}\cdot \dfrac{r^{n}-1}{r-1}](https://tex.z-dn.net/?f=S_%7Bn%7D%3Da_%7B1%7D%5Ccdot%20%5Cdfrac%7Br%5E%7Bn%7D-1%7D%7Br-1%7D)
Filling in your numbers, this is
![S_{180}=300(1+\frac{0.017}{12})\left( \dfrac{(1+\frac{0.017}{12})^{180}}{(\frac{0.017}{12})} \right) \approx 61547](https://tex.z-dn.net/?f=S_%7B180%7D%3D300%281%2B%5Cfrac%7B0.017%7D%7B12%7D%29%5Cleft%28%20%5Cdfrac%7B%281%2B%5Cfrac%7B0.017%7D%7B12%7D%29%5E%7B180%7D%7D%7B%28%5Cfrac%7B0.017%7D%7B12%7D%29%7D%20%5Cright%29%20%5Capprox%2061547)
If Jerry's contributions and withdrawal are at the end of the month, this balance is reduced by 1 month's interest, so is $61,460.
_____
We suppose the expected choice is $61,960. This supposition comes from the fact that a handwritten 4 is often confused with a handwritten 9. The usual simple calculation of future value uses end-of-the-month contributions by default. (a₁ = 300)
Answer:
z = -11
Step-by-step explanation:
ALL U HAVE TO DO IS THE OPPOSITE OF ³
WHICH IS ²
I think it’s the last answer.
Step-by-step explanation:
-4 is answer to this problem