A scholarship fund is accumulated by deposits of $400$ at the end of each year. The fund is to be used to pay out one annual scholarship of $2000$ in perpetuity, with the first scholarship being paid out one year after the last deposit. Assume $i=0.08$. Find the minimum number of deposits which must be made in order to support such a fund.
I am not quite sure how I started doing this question but its probably wrong anyways. I was trying to use a similar question from the textbook to figure it out. But what I got was $$\frac{2000}{1.08-1}=25 000$$ and then using $400 \ddot s _n \le 25000 \lt 400 \ddot s_{n+1}$ which then by simplifying the equation I got $\ddot s _n \le 62.5 \lt \ddot s _{n+1}$. I have no clue if any of this is right. Or how to get the right answer. I know the answer is supposed to be $24$ but I don’t know how.
We must have $$ 400\,s_{\overline{n}|i}=2000\,a_{\overline{\infty}|i} $$ that is $\frac{(1+i)^n-1}{i}=\frac{5}{i}$ and then $(1+i)^n=6$ and finally $$ n=\frac {\log 6}{\log(1+i) }\approx 23.28 $$
So the minimum number of years is 24.