Annuity: interest conversion more frequent than payment

Question

Gus deposits 25 into a bank account at the beginning of each 3-year period for 18 years (i.e. there is no deposit in year 18). The account credits interest at an annual effective rate of interest of i. The accumulated amount in the account after 18 years is X, which is four times as large as the accumulated amount in the account after 9 years (excluding the deposit made at time t = 9). Calculate X.

Correct answer: 652.23

My work:

$n$=total number of conversion periods for annuity term = 18, $k$ = number of conversion periods per payment period = 3

Then using equations for level annuities with payment period greater than interest conversion period, we solve $$ 25\frac{s_{\bar{18|}}}{a_{\bar{3|}}} = 4 \left[25\frac{s_{\bar{6|}}}{a_{\bar{3|}}} + 25\right] $$

But I get stuck on solving for the interest rate, I get a polynomial with terms to powers 21, 9, and 6.

Answer 1

$$X = 25(1+i)^{18} + 25(1+i)^{15} + \cdots + 25(1+i)^3 = 25 \require{enclose}\ddot s_{\enclose{actuarial}{6} j},$$ where $j = (1+i)^3 - 1$ is the effective three-year periodic rate of interest. We are also given $$\frac{X}{4} = 25(1+i)^{9} + 25(1+i)^{6} + 25(1+i)^{3} = 25 \require{enclose}\ddot s_{\enclose{actuarial}{3} j},$$ where again $j$ is defined as above. Therefore, $$4 \require{enclose}\ddot s_{\enclose{actuarial}{3} j} = \ddot s_{\enclose{actuarial}{6} j},$$ or equivalently, $$4 (1+j) \frac{(1+j)^3 - 1}{j} = (1+j) \frac{(1+j)^6 - 1}{j},$$ hence $4((1+j)^3 - 1) = (1+j)^6 - 1$, or $(1+j)^3 = 3$. We ignore the trivial root $j = 0$. Then to compute $X$, we simply substitute: $$X = 25 (1+j)\frac{(1+j)^6 - 1}{j} = 25 \cdot 3^{1/3} \frac{3^2 - 1}{3^{1/3} - 1} \approx 652.233.$$

Answer 2

After 9 years, the balance is $B=25\left((1+i)^9+(1+i)^6+(1+i)^3\right)$. After 18 years, the balance is $B'=25\left((1+i)^{18}+(1+i)^{15}+(1+i)^{12}+(1+i)^9+(1+i)^6+(1+i)^3\right)=B\left(1+(1+i)^9\right)$. We are given $B'=4B$ so $1+(1+i)^9=4$ hence $i=\sqrt[9]{3}-1$, and so plugging this into $B'$ gives the answer.