A mixture of AP and GP

Question

A battery loses $10$ mAh of after every hour when in use. The same battery loses $1\%$ of its current amount of charge every hour when not in use.

Suppose that the battery is fully charged with $1000$ mAh. The battery is used for an hour, left unused for the next hour, use for the consequent hour, and left unused again for the consequent hour.

How much charge does the batter have left after $2n$ hours, where $n$ is a positive integer?

I came up with a recurrence relation to describe the amount of charge left after $2n$ hours:

$$A_n = (A_{n-1} - 10)\cdot0.99\\=0.99A_{n-1} - 9.9$$

And then I rearrange to get

$$A_n -0.99A_{n-1} = -9.9$$

Now, I suppose that I would have to apply some sort of summation to both sides such that the LHS experiences a whole lot of summation, to leave $A_n$. However, I could not think of anything.

Is my method sound? Also, what summation should I apply? This came out in today's test and I ended up confusing myself with my workings (3-hr long papers are tiring).

Answer 1

Your working is sound, here is what you do, to sum a mixture of A.P. and G.P.

Given a sum of the form,

$$\underbrace{a+(a+d)k+((a+d)k+d)k+\ldots+((a+d)k+\cdots+d)k}_{\text {n terms}}$$

We want to obtain a closed form solution.

$$\underbrace{a+ak+ak^2+\cdots+ak^n}_{\text{Geometric Progression} = S_1} +dk+dk^2+dk+dk^3+dk^2+dk+\cdots+dk^n+dk^{n-1}+\cdots dk$$

$$S_1+ ndk+(n-1)dk^2+(n-2)dk^3+\cdots+dk^n$$

$$S_1+\underbrace{n(dk+dk^2+\cdots+dk^n)}_{\text{Geometric Progression}=S_2}-d(k^2+2k^3+3k^4+\cdots+(n-1)k^n)$$

Can you do it now?

I think there might be a better solution though.

Answer 2

Charge consumed in $2n$ hours = $$d+0.01(a-d)+d+0.01(a-2d) + \dots + d + 0.01(a-nd) = nd+0.01(na-\frac{n(n+1)}{2}d)$$ where $d=10$ and $a=1000$.