How many days will I need to finish a book if I read more pages every day?

Question

Lets say a book has 300 pages and I read 10 pages a day + 5% more pages every day (10 x (1 + 0.05)). So the first day I read 10 pages, 2nd day I read 10.5, 3rd day I read 11,025 pages.. and so on. How many days will I need to finish a book of 300 pages?

Answer 1

On the first day you read $10$ pages. On the second day you're going read $10(1+0.05)=10\times1.05=10.5$ pages. On the third day you will read $10\times1.05^{2}=11.025$ pages.

So at the end of the third day you've read a total of $10\times(1+1.05+1.05^{2})=31.525$ pages.

You can see that on at the end of the $N^{th}$ day, the total pages read will be $10\times(1+1.05+1.05^{2}+...+1.05^{N-1}).$

Now $\sum_{k=0}^{n-1}ar^k=a\frac{1-r^{n}}{1-r}$ is the sum of the first $n$ terms of a geometric series.

So using this information, on the $N^{th}$ day you will have read a total of

$$10\times(1+1.05+1.05^{2}+...+1.05^{N-1})=\sum_{k=0}^{N-1}10(1.05)^{k}=10\frac{1-(1.05)^N}{1-1.05}=200((1.05)^{N}-1) \space\space \text{pages}$$

You want to read $300$ pages, so solving $$200((1.05)^{N}-1)=300$$

$$(1.05)^{N}-1=1.5$$

$$N\ln(1.05)=\ln(2.5)$$

Thus $N=\frac{\ln(2.5)}{\ln(1.05)}=18.7802...$ So you need about $19$ days to finish a book of $300$ pages.