How many monthly deposits of 100 each and what final deposit one month later will be necessary to accumulate 3000 if interest is at 9% compounded quarterly?
Monthly deposit?
Final deposit?
To calculate the monthly deposit, I first calculated the monthly rate equivalent to the nominal interest rate compounded quarterly:
(1 + i)^12 = (1 + (0.09/4))^4
1 + i = (1.0225)^(4/12)
i = (1.0225)^(1/3) - 1
i = 0.007444443
Then I calculated n periods for which there are monthly deposits:
100(1 + 0.007444443)^n = 3000
n log (1.007444443) = log 30
n = 458.5758454 months or 38.21465378 years
How to calculate the monthly deposits and the final deposit?
Thanks for any help!
The first deposit earns interest for $n$ months. The second deposit earns interest for $n-1$ months, and the $n$-th deposit earns no interest. That gives us
$$ 100(1+i)^{n} + 100(1+i)^{n-1} + \cdots + 100(1+i) + 100 = 3000 \enspace. $$
We rewrite the sum on the left as
$$ 100\, \sum_{0 \leq k \leq n} (1+i)^k = 3000 \enspace, $$
and we use the formula for the partial sum of a geometric series:
$$ 100\, \frac{(1+i)^{n+1} - 1}{1+i-1} = 3000 \enspace. $$
A little algebra sees us through:
$$ n = \frac{\log(1 + 30i)}{\log(1+i)} - 1 = 26.18\enspace. $$
After $26$ months we are only left with the final payment. The value accumulated in those 26 months is computed by the same formula we used to derive $n$, except that now $n=26$:
$$ 100\, \frac{(1+i)^{27} - 1}{i} = 2978.26 \enspace. $$
The amount left to deposit on top of the usual $100$ to get to $3000$ is therefore $21.74$.