I have spent several hours trying to solve this problem. $A$ and $B$ both open up new bank accounts at time $0$. The principle for $A$ (the amount deposited at $t=0$) is $100$. The principle of $B$ is $50$ (the amount deposited at $t=0$). Each account earns an annual discount rate of $d$. The amount of interest earned in $A$ during the 11th period is equal to $X$. The amount of interest earned in $B$ during the 17th period is equal to $X$. Calculate X.
Given that we are dealing with effective discount rate, for $A$ and $B$ we have $a(t)^{-1} = (1-d)^t$ clearly the amount of interest earned is $(1-d)^t$ for both $A$ and $B$. This means that $(1-d)^{11}=(1-d)^{17}$ Is this the correct set up? if not why? Given that the principles are $100, 50$ we have:
$\frac{100}{(1-d)^{11}}=\frac{50}{(1-d)^{17}}$ that is if we are setting the amount functions equal to each other during the $17$ and $11$ period. I thought this would translate to $50(1-d)^{11} = 100(1-d)^{17}$ then substituting $X$ for $(1-d)^{17}$ and $(1-d)^{11}$ we have something that makes no sense $50X=100X$ which gives $X=0$ after solving. If I don't substitute for $X$ I have $50(1-d)^{11}=100(1-d)^{17} \to 50=100(1-d)^6$ which translates to $\frac{1}{2}=(1-d)^{6}$ which gives a decimal less than one. The answer is $38.88$ I have no idea where I am going wrong. What am I missing? Am I even any where close on my thought processes?
The amount of interest earned in A during the 11th period is equal to $100\cdot (q^{11}-q^{10})$, where $q=1+i$ and $i$ is the interest. Equivalently for $B$. Thus the equation is
$$100\cdot (q^{11}-q^{10})=50\cdot (q^{17}-q^{16})$$
Dividing the equation by $50\cdot q^{10}$.
$$2\cdot (q-1)=(q^{7}-q^{6})$$
The (approximate) solution is $q=1.12246$
Thus $X=50\cdot (1.12246^{17}-1.12246^{16})=38.87749...\approx 38.88$