# Combined arithmetic and geometric progression problem

Tom fertilizes his plant once a week. Every week the amount of total fertilizers reduces by 25%. Let's say that every Saturday Tom adds 10 grams of fertilizers to his plant.

Write an equation to calculate the amount of fertilizers after every fertilization.

So first I realized that $$a_{n} = a_{n-1} \cdot q + d$$ which in this case is

$$a_{n} = 0.75 a_{n-1} + 10$$

a.k.a

$$(((10 \cdot 0.75 + 10) \cdot 0.75 + 10) \cdot 0.75 + 10) \cdots$$

I have also noted that the amount of reduction of fertilizers is going to be 1 less than the amount of addition of fertilizers.

But how to write this as an equation? I haven't progressed for a while now with this problem and I hope you can help me with an explanation behind your thoughts!

Let $a_m=b_m+C$

$\implies b_n+C=0.75[b_{n-1}+C]+10$

$\iff b_n+C=0.75b_{n-1}+0.75C+10$

Set $C=0.75C+10\iff C=?$ to get $\iff b_n=0.75b_{n-1}=\cdots(0.75)^rb_{n-r}$ where $r$ is any integer.

You must have an initial condition, I presume?

In effect we find, $a_n-40=0.75[a_{n-1}-40]$

$$\begin{cases} a_0=k\\a_n=a_{n-1}q+d\end{cases}$$

Is solved by

$\begin{cases} \text{if }q\neq1,\ a_n=q^nk+\frac{q^n-1}{q-1}d\\\ \text{if }q=1,\ a_n=k+nd\end{cases}$

In your case, $q=0.75,\ d=10$ and $k$ being the initial quantity of fertilizer (which you didn't state).

Note that, after a long period of time (a.k.a. for big values of $n$), since $|q|<1$, $a_n\sim\frac{d}{1-q}=\frac{10}{0.25}=40$, idependently of the initial dose of fertilizer.

For $q\neq1$, let's show that $q^nk+\frac{q^n-1}{q-1}d$ is the solution of $$\begin{cases} a_0=k\\a_n=a_{n-1}q+d\end{cases}$$

(if we use the convention $0^0:=1$)

Indeed, let $m$ the smallest integer such that $a_m\neq q^mk+\frac{q^m-1}{q-1}d$.

Then, $m>0$, since $a_0=k$ and $q^0k+\frac{q^0-1}{q-1}=k+\frac{0}{q-1}$.

But $m>0\rightarrow m=t+1$, with $t\geq0$.

By minimality of $m$, $a_t=q^tk+\frac{q^t-1}{q-1}d$. By definition, $$a_m=a_tq+d=q^{t+1}k+\frac{q^{t+1}-q}{q-1}d+d=q^{m}k+\frac{q^m-1}{q-1}d$$

• Then $k=0$, hence $a_n=40\cdot(1-0.75^n)$. Ok, I'll add a proof (the explanation would be that I know the formula for $\sum_0^n\alpha^s$)