Hi,
I've got a question regarding superannuation and loan repayments in the applications of series topic.
For supperannuation what are the formulas that's needed and the process of finding how much an amount has grown from one year to another year and the total investment overall.
For loan repayments what are the formulas thats needed and the process to it?
I just find these two a really long process that I kind of have trouble revising.
Here's an example of questions
Thanks 
 is putting a spin on things by focusing on}\\ \textbf{each single investment one at a time.})
This is in contrast to the usual recursive method, where we have a running total of what goes on but don't really care about what happens to each individual investment.
^{30}.)
The other part is just the old usual recursive approach.
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Personal opinion: In the rare case this happens, I like to do the usual thing with whichever is more frequent. Here, the compounding frequency is just monthly, whereas the payments are made annually, so I analyse the whole thing on a monthly basis.
The recursive approach should be understood by trying to read out loud what is going on. Here, between the jump from the beginning to the end of the first month, the only thing that occurs is that it picks up one month's worth of interest. So at this stage, we still require only the compound interest formula.
\begin{align*}A_1 &= A_0 (1.01)\\ \therefore A_1 &= 50000(1.01) \end{align*}
Between the end of the first month and the end of the second month, the same thing happens.
\begin{align*}A_2 &= A_1(1.01)\\ \therefore A_2 &= 50000(1.01)^2\end{align*}
And it should be clear that similarly, \(A_3 =50000(1.01)^3\). This repeats itself all the way up to \(A_11\).
\[ A_{11} = 50000(1.01)^{11} \]
And then after the 12th month, after the money picks up interest yet again, we make our first payment. Recall that the payment we make is \(M\). So we have two things going on here now; the interest, followed by payment.
\begin{align*}A_{12} &= A_{11}(1.01) - M\\ &= 50000(1.01)^{12} - M. \end{align*}
^{24} - M(1.01)^{12} - M\end{align*}\\ \text{which you can just set equal to 0.}\\ \text{Note that for this particular problem no geometric progression was really necessary;}\\ \text{there weren't that many terms in our series to begin with.})
Remark: iv) is nothing fancy; they just want you to calculate what the "average interest across the two years" was, which you can do by comparing how much you had to pay off (i.e. \(2M\)) to the initial amount.
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