I state that I am italian, so, if there are some mistake in my questions say it to me, and I correct as soon as possible. I've known this forum in an italian forum about molecular biology, where I was asking a problem about financial mathematics. Here is the problem: "For a loan of € 4,700 received today, to settle in a year, you are paid 3 monthly installments of € 800 and then 3 monthly installments of € 850. How wise was the case the loan? "The installments are in arrears.
Ok, the problem may appear easy, and I'm thinking about it since one week, but I haven't find solution. My thoughts are: -I've tried with the matimatical formula:$ M=R[k+r((k-1)/2)] $ and $ r=(M/R-k)/((k-1)/2) $ but the installments are not costant and regular. In fact, are yes constant in the time (bimestral) but not constant for the amount (800 and 850). So the formula doesen't work. -I've checked if I need to use tha financial table, but (I think) are not to use. -I've serched on another books and on Internet if there are other formulas orthoughts to use but I haven't found nothing.
Could someone help me please? I'm very interesting about the thoughts for arrive at the result :)
Sorry for the english.
Hmmm. Okay so you get €4700 and you pay €800, €800, €800, €850, €850, €850, one payment every two months. These numbers count out to €4950.
The bad news is I don't a fixed formula is going to pull this off.
We'll assume compound interest is involved; this makes the calculations a little complicated so I went after it in Excel with Goal Seek. This gave me an annual interest rate of about 8.9%. We can, however, take it on a little bit with more serious tools.
$$P_0 = 4700$$ $$P_1 = rP_0 - 800$$ $$P_2 = rP_1 - 800$$ $$P_3 = rP_2 - 800$$ $$P_4 = rP_3 - 850$$ $$P_5 = rP_4 - 850$$ $$P_6 = rP_5 - 850 = 0$$ $$P_5 = \frac{850}{r}$$ $$P_4 = \frac{850}{r^2}+\frac{850}{r}$$ $$P_3 = \frac{850}{r^3}+\frac{850}{r^2}+\frac{850}{r}$$ ... and so on, giving... $$4700 = \frac{850}{r^6}+\frac{850}{r^5}+\frac{850}{r^4}+\frac{800}{r^3}+\frac{800}{r^2}+\frac{800}{r}$$
At which point it's much nicer to think in terms of $q=\frac{1}{r}$.
$$4700 = 850q^6+850q^5+850q^4+800q^3+800q^2+800q$$
Reducible:
$$17q^6+17q^5+17q^4+16q^3+16q^2+16q-94=0$$
This is kind of big, and unfortunately all the candidates for rational roots fail. We do know there's only one positive real root, and we can find it by Newton's Method: $q \approx 0.985401$. This gives $r \approx 1.014814$, or about $1.48\%$. But this is the rate for one of the six terms; to find the annualized rate, we multiply that by 6, giving $8.88\%$, so we've confirmed what Excel gave me.