In the sequence 1, 3, 5, 7, ..., k, ..., 57, the sum of the numbers up to k, excluding k, is equal to the sum of the numbers from k up to 57, also excluding k. What is k?
Show Hint
Instead of first locating the exact position of $k$ in the list, try expressing "sum of terms before $k$" directly as a function of $k$, using the fact that those terms form a shorter arithmetic progression with the same common difference. Combine this with the fact that the two given sums must each equal half of everything except $k$.
Step 1: The sum of the first \(n\) odd numbers is always \(n^2\). Here \(57\) is the \(29\)th odd number, so the total sum is \(29^2=841\). Step 2: If \(k\) is the \(m\)-th odd number, the terms before it sum to \((m-1)^2\), and this must equal half of everything except \(k\): \[ (m-1)^2=\frac{841-k}{2}, \quad k=2m-1. \] Step 3: Substituting and simplifying gives \(m^2-m-420=0\), which solves to \(m=21\). Step 4: So \(k\) is the \(21\)st odd number: \[ \boxed{k=2(21)-1=41} \]