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Consider an investment with nonzero interest rate $i$. If $i_5$ is equal to $i_{10}$, show that interest is not computed using simple interest.

Answer is If $i$ is a simple interest rate, then $i_5=i_{10}$ implies $i=0$

Workings:

$i_5=\frac{A(5)}{A(4)}-1$

$i_{10}=\frac{A(10)}{A(9)}-1$

If equal, then by cancelling ($-1$) on both sides,

$\frac{A(5)}{A(4)}=\frac{A(10)}{A(9)}$

$A(0) \frac{a(5)}{a(4)}=A(0) \frac{a(10)}{a(9)}$

$\frac{a(5)}{a(4)}=\frac{a(10)}{a(9)}$

From here, I am stuck.

Moody Blue
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2 Answers2

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For a proof by contradiction, assume simple interest:

$A(5)=A(0)(1+5i)$

$A(4)=A(0)(1+4i)$

$A(10)=A(0)(1+10i)$

$A(9)=A(0)(1+9i)$

$i_5=\frac{A(5)}{A(4)}-1$

$i_5=\frac{A(0)(1+5i)}{A(0)(1+4i)}-1=i_5=\frac{(1+5i)}{(1+4i)}-1$

$i_10=\frac{A(0)(1+10i)}{A(0)(1+9i)}-1=i_{10}=\frac{(1+10i)}{(1+9i)}-1$

Solving for $i_5=i_{10},

$\frac{(1+5i)}{(1+4i)}-1=\frac{(1+10i)}{(1+9i)}-1$

$5i^2=0$

$i=0$

For equality to hold, $i=0$, contradicting the conditions in the question, hence simple interest will not be applied for $i>0$

Tosh
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Assume that the interest is computed using simple interest. On the one hand, $i_5$ is given by: $$i_5 = P\ i\ t_5,$$ $P$ being the initial investment and $t_5$ the time since then. On the other hand, analogously: $$i_{10} = P\ i\ t_{10}.$$ Since you are told that $i_5 = i_{10}$, you can deduce that $$P\ i\ t_5 = P\ i\ t_{10}$$ Obviously, $t_5<t_{10}$ and $P\ne0$ (I suppose you really have invested money...), so the only posibility for the equality to be true is that $i=0$, which is not true. Therefore, our first assumption (simple interest) is not true neither.

AugSB
  • 5,007