# Thread: Sequences and Series

1. ## Sequences and Series

Hey, i need help with this question, could you please lend a helping hand
This is the question:

Jen plans to invest $10 000. She plans to travel through EUrope when she graduates from university in five years. Jen deposits the 10 000 today and leaves it in the bank earning an interest rate of 6.5%/a compounded semi-annually. a) If she withdraws$8 000 at the end of 5 years for her trip to Europe, how much will she still have in her account? Show all calculations.
b) If this amount (your answer from part a) is left in the account at the same interest rate, how much would she have in her account to put towards a car when she returns from Europe a year later?

2. This really isn't a question on sequences & series, but on exponential functions. I probably would have put this question in the Pre-Calculus subforum.
Originally Posted by dragasma
Jen plans to invest $10 000. She plans to travel through EUrope when she graduates from university in five years. Jen deposits the 10 000 today and leaves it in the bank earning an interest rate of 6.5%/a compounded semi-annually. a) If she withdraws$8 000 at the end of 5 years for her trip to Europe, how much will she still have in her account? Show all calculations.
b) If this amount (your answer from part a) is left in the account at the same interest rate, how much would she have in her account to put towards a car when she returns from Europe a year later?
Use the equation
$\displaystyle A = P\left(1 + \frac{r}{k}\right)^{kt}$
where:
P = initial investment
r = annual interest rate
k = number of times the interest is compounded per year
t = time in years
A = amount after time t

In your case, P = 10,000, r = 0.065, k = 2, and t = 5:
\displaystyle \begin{aligned} A &= 10000\left(1 + \frac{0.065}{2}\right)^{2(5)} \\ &= 10000\left(1.0325\right)^{10} \\ &\approx 13768.94 \end{aligned}
After 5 years, after withdrawing $8,000.00 from her account, she will have$5,768.94 left.

For part b, P is now 5768.94:
\displaystyle \begin{aligned} A &= 5768.94\left(1 + \frac{0.065}{2}\right)^{2(1)} \\ &= 5768.94\left(1.0325\right)^{2} \\ &\approx 6150.01 \end{aligned}
Jen will have \$6,150.01 in her account when she returns from Europe a year later.

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