# AoPS Wiki talk:Problem of the Day/June 15, 2011

## Solution

### Solution 1

We can solve this problem by a bit of trial and error.

We can guess she rode $5$ days and we get $7+10+13+16+19=(13)(5)=65$ since the mean is clearly $13$ and there are $5$ terms.

That's a bit too small.

We can add $22$ to $65$ and get $87$. That's still too small.

Now, we add $25$ to get $112$, which is the total we want.

We started with $5$ days and added $2$ more, so there are $7$ days in total. We can add $7-1=6$ days to her starting day, Monday, to find her final day. Our answer is thus $\boxed{Sunday}$

### Solution 2

On the first day, Jenny rode $7$ miles. On the second day, she rode $7+3=10$ miles. On the third day, she rode $10+3=13$ miles.

This is the sequence $7,10,13,...$ which is an arithmetic sequence: first term $7$, common difference $3$.

We are trying to find the number of terms $n$ such that the $n\text{th}$ partial sum of the sequence is $112$.

The formula for the sum of a partial sequence is $\frac{n}{2}[2a+(n-1)d]$, where $a$ is the first term, $n$ is the number of terms, and $d$ is the common difference. (Try to derive it!)

Let $a=7$ and $d=3.$ Then we have: $\frac{n}{2}[14+3(n-1)]=112$ $n[14+3(n-1)]=224$ $14n+3n(n-1)=224$ $14n+3n^2-3n=224$ $3n^2+11n-224=0$ $(n-7)(3n+32)=0$

The second root is not an integer, so the workout lasted for $n=7$ days. The $7\text{th}$ day after Monday is $\boxed{\text{Sunday}}$.