Question about Interest

Question

An accumulation function is of the form $a(t) = at^2 + bt + c$. Rick deposited $\$10,000$ in an account that is subject to this accumulation function. After $7$ years, Rick has $\$12,870$ in his account. Leah deposited $\$25,000$ in another account that is subject to the same accumulation function. After 15 years, Leah has $\$49,375$ in her account. Find Leah's effective rate of interest for the third year.

Answer 1

$a(7) = 49a + 7b +c$

$a(15) = 225a+15b + c$

$a(0) = c = 1$

$a(7) = \frac{12870}{10000} = 49a+7b+1=1.287$

$a(15) =\frac{49375}{25000} = 225a+15b+1 = 1.975$

$49a + 7b = 0.287$

$225a +15b = 0.975$

Solving the simultaneous equations we get $a = 0.003$ and $b = 0.02$

Now for the third year, we simply calculate

$9a+3b+1 = 1.087$

$4a+2b+1 = 1.052$

Thus $a(3) =1.087$

$a(2) =1.052$

$ i = \frac{1.087}{1.052}-1 = 3.296$% is the effective interest rate for the third year.

Answer 2

Observe that$$ \frac{A(t)}{A(0)}=a(t)=at^2+bt+c=at^2+bt+1\quad\text{because }c=a(0)=\frac{A(0)}{A(0)}=1 $$ The annual effective rate of interest for year $t$, which we denote by $i(t)$, is the ratio of the amount of interest earned in a year, from time $t−1$ to time $t$, to the accumulated amount at the beginning of the year $$ i(t)=\frac{A(t)-A(t-1)}{A(t-1)}=\frac{a(t)-a(t-1)}{a(t-1)}=\frac{a(t)}{a(t-1)}-1 $$ For Rick $A(0)=\$ 10,000$ and $A(7)=\$ 12,870$ so that $a(7)=1.285$; for Leah $A(0)=\$ 25,000$ and $A(15)=\$ 49,375$ so that $a(15)=1.975$.

Then $$\begin{align} a(7)-1=0.285&=a\cdot 7^2+b\cdot 7\\ a(15)-1=0.975&=a\cdot 15^2+b\cdot 15 \end{align} $$ that is $a≈0.00303571$, $b≈0.0194643$.

The annual effective rate of interest for year $t=3$ is $$ i(3)=\frac{a(3)}{a(2)}-1=3.296\% $$