What would the equation for this word problem be?

Question

I don’t even know if there is an equation for this kind of problem, but:

“A stock starts with a value of $100 per share. On the first day, the stock’s value increases by 12.0%. The next day, the stock’s value decreases by 12.0%. On the third day, the stock’s value increases by 12.0% again. On the fourth day, the stock’s value decreases by 12.0% again. This back and forth continues for 178 days so that, in the end, the stock has gone through 89 up and down cycles. How much is each share of this stock worth at the end of day 178?”

I’m eager to learn how it is that this word problem can be figured out. Thank you!

Answer 1

A $12\%$ increase changes the value to $$ \$100 \times 1.12 = \$112. $$ Then a $12\%$ decrease makes it $$ \$100 \times 1.12 \times 0.88 = \$100 \times 0.9856 =\$98.56. $$ Each two day cycle reduces the value to $98.56\%$ of what it was. After $89$ cycles (half a year) it's worth $$ \$100 \times 0.9856^{89} = \$27.50. $$

Answer 2

Let's call $S(n)$ the price of the share after $n$ days. Then, observe the following pattern that emerges from the behaviour of the first few days $$S(0)=100,$$ $$S(1)=100\times1.12,$$ $$S(2)=100\times 1.12\times 0.88,$$ $$S(3)=100\times1.12\times0.88\times1.12=100\times1.12^2\times0.88,$$ $$S(4)=100\times1.12^2\times0.88\times0.88=100\times1.12^2\times0.88^2.$$ We may generalise at this point. For odd $n$ we get $$S(2n+1)=100\times 1.12^n\times0.88^{n+1},$$ and for even $n$ $$S(2n)=100\times1.12^{n}\times0.88^n.$$ So, for $178$, we have $$S(178)=100\times 1.12^{89}\times 0.88^{89}.$$

Answer 3

By calling $u_n$ the value per share at day $n$, When you increase by $12%$, you multiply your value by $1+0.12=p$ When you decrease by $12%$, you multiply your value by $1-0.12=q$ $u_1 = p u_0$, $u_2 = q u_1$. In general, $u_{2n+1} = p u_{2n}$ and $u_{2n} = q u_{2n-1}$. $u$ will have the form: $u_n = q^{f(n)} p^{g(n)} u_0$ It will depend on the parity: If $n=2m$, there are $m$ $p$ and $m$ $q$ applied to $u_0$ as in the example : $u_2 = pqu_0$. If $n=2m+1$, there are $m+1$ $p$ and $m$ $q$ applied to $u_0$. $u_{2n} = (pq)^n u_0$ and $u_{2n+1} = p (pq)^n u_0$.