Annuity payments $2n = n$ I'm having trouble with this question:
For an annuity of $3n$ payments of equal amount at periodic interest rate $i$, it is found that one period before the first payment the present value of the first $n$ payments is equal to the present value of the final $2n$ payments.  What is the value of $v^n$?
My thoughts are to make $a_n = a_{2n}$, but I don't know if that is correct
I get $(1-v^n)/i = (1-v^{2n})/i$.
 A: At the indicated valuation point, the present value of the first $n$ payments is given by $$K a_{\overline{n}\rceil i} = K \frac{1 - v^n}{i}$$ where $K$ is the level payment amount per period, $v = (1+i)^{-1}$ is the present value discount factor, and $i$ is the effective periodic interest rate.  The present value of the final $2n$ payments is $$K \,_{n|} a_{\overline{2n}\rceil i} = K v^n \frac{1 - v^{2n}}{i},$$ because the final $2n$ payments is deferred by $n$ payments.  To see this more clearly, write out the cash flow:  $$PV = (Kv + Kv^2 + \cdots + Kv^n) + (Kv^{n+1} + Kv^{n+2} + \cdots + Kv^{3n}).$$  The first $n$ payments and the final $2n$ payments are grouped together.  Then we see that the second group can be factored as $$Kv^n (v + v^2 + \cdots + v^{2n}).$$  Therefore, the equation to be solved is $$1-v^n = v^n(1-v^{2n}).$$  This is a cubic polynomial in $v^n$:  $$(v^n)^3 - 2(v^n) + 1 = 0.$$  It is an easy exercise to show that the only root for which the resultant interest rate and number of payments makes sense is when $v^n = (\sqrt{5}-1)/2$.  However, without further information about the value of $i$, it is not possible to furnish a unique integer solution for $n$.  For example, if $i \approx 0.173985$, then $n = 3$, but if $i \approx 0.0134567$, then $n = 36$.
