I am working on the following problem that involves annuity.
On January 1, 1999, Biff purchases an annuity for 54700 dollars. The annuity makes annual payments of the form X, 2X, X, 2X, … with the first payment coming on January 1, 2000, and the final payment coming on January 1, 2040. Assuming an effective rate of 8.5 percent, what is X?
I started solving this problem like this: $$ X\cdot(s_\overline{41|}+s_\overline{20|})=54,700 $$ $$ X\cdot(\frac{1.085^{41}-1}{0.085}+\frac{1.085^{20}-1}{0.085})=54,700 $$ Then, solving for X I got $$ X=147.761 $$
I also tried to solve it like this:
$$
X\cdot(s_\overline{21|}+2s_\overline{20|})=54,700
$$
$$
X=364.077
$$
Then I tried to use $a_\overline{n|}$ instead of $s_\overline{n|}$ and $s_\overline{40|}$ instead of $s_\overline{41|}$.
But WebWork says neither of the answers that I got is correct.
Can I have some help?
Thank you.
I assume that payments are made once a year (not: monthly!) on 01. January.
Observe the first payment in the year 2000 is X, and the last payment in the year 2040 is again X.
Let $n$ be the number of two-year periods. On 2. January 2000, you have $a_0 = 54,700 \cdot 1.085 - X$. Then you have the recursion $a_{n+1} = (a_n \cdot 1.085 - 2 X) \cdot 1.085 - X $. This recursion occurs 20 times. Let's compute $a_{20}$ with $a_{n+1} = a_n \cdot 1.085^2 - (2 \cdot 1.085+1) X $ which gives
$$ a_{20} = - (2 \cdot 1.085+1) X + 1.085^2 \cdot a_{19} = \\ - (2 \cdot 1.085+1) X (1 + 1.085^2 + 1.085^4 + \cdots + 1.085^{2\cdot 19}) + 1.085^{2\cdot20} a_0 = \\ - (2 \cdot 1.085+1) X \frac{1-1.085^{2\cdot 20}}{1 - 1.085^2}+ 1.085^{2\cdot20} a_0 $$ and finally $$ 0 = a_{20} = - (2 \cdot 1.085+1) X \frac{1-1.085^{2\cdot 20}}{1 - 1.085^2}+ 1.085^{2\cdot20} (54,700 \cdot 1.085 - X) $$ or $$ X \left[ (2 \cdot 1.085+1) \frac{1-1.085^{40}}{1 - 1.085^2} + 1.085^{40} \right] = 1.085^{41} \cdot 54,700 $$ which solves for $X = 3260.53$
For comparison:
without any interest, we would have $61 \cdot X = 54,700$ or $X \simeq 897$.
with the given interest rate, paying the full amount $61 \cdot X$ on 01. January 2040 gives $61 \cdot X = 54,700 \cdot 1.085^{41}$ or $X \simeq 25,426$.