Question

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?

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 $k$.

Answer 1

Correct Answer: 41

Approach: The numbers are consecutive odd numbers, and the sum of the first \(n\) odd numbers is the clean perfect square \(n^2\). That single fact turns the whole problem into a tiny quadratic.

Step 1: List the sequence \(1, 3, 5, \ldots, 57\). General term \(a_n = 2n-1\). Since \(2N-1 = 57\), there are \(N = 29\) terms, and the total sum is \(1+3+\cdots+57 = 29^2 = 841\).

Step 2: Let \(k\) be the \(m\)-th term, so the terms before \(k\) are the first \(m-1\) odd numbers, summing to \((m-1)^2\). The terms after \(k\) sum to (total) minus (first \(m\) terms) \(= 841 - m^2\).

Step 3: The condition "sum before \(k\) = sum after \(k\)" gives \[ (m-1)^2 = 841 - m^2. \] Expand: \(m^2 - 2m + 1 = 841 - m^2\), so \(2m^2 - 2m - 840 = 0\), i.e. \(m^2 - m - 420 = 0\).

Step 4: Factor: \((m-21)(m+20)=0\), so \(m = 21\) (reject \(m=-20\)).

Step 5: \(k = a_{21} = 2(21)-1 = 41.\)

Answer: \(\boxed{k = 41}\).

Answer 2

Approach: Apply the arithmetic-series sum formula directly to the "before" and "after" portions, without relying on any odd-number shortcut.

The full sequence is \(1,3,5,\ldots,57\), common difference \(2\), so if \(k\) is the \(m\)-th term then \(k=2m-1\), and there are \(29\) terms in total (since \(2(29)-1=57\)).

Terms before \(k\): these are \(1,3,\ldots,(k-2)\), which is \((m-1)\) terms with first term \(1\) and last term \(k-2\). Their sum is \[ S_{\text{before}}=\frac{(m-1)}{2}\big(1+(k-2)\big)=\frac{(m-1)(2m-2)}{2}=(m-1)^2. \]
Terms after \(k\): these are \((k+2),\ldots,57\), which is \((29-m)\) terms with first term \(k+2\) and last term \(57\). Their sum is \[ S_{\text{after}}=\frac{(29-m)}{2}\big((k+2)+57\big). \]
Setting \(S_{\text{before}}=S_{\text{after}}\) and substituting \(k=2m-1\) gives, after simplification, \(m^2-m-420=0\), so \(m=21\) and \[ k=2(21)-1=\boxed{41} \]

Answer 3

Concept:

• The total sum of the sequence stays fixed; removing $k$ from it leaves "sum before $k$" plus "sum after $k$" together equal to the total minus $k$.
• The given condition states these two parts are equal to each other, so each one equals exactly half of (total $- k$), without ever needing a separate expression for the "after" portion.
• The terms before $k$ form a shorter arithmetic progression with the same common difference, so their sum can be written directly in terms of $k$, without first finding the position of $k$ in the list.

Step 1: Find the total sum of the sequence.
$1, 3, 5, \dots, 57$ is an AP with first term $1$, common difference $2$. Number of terms: $57 = 1 + (n-1)2 \Rightarrow n = 29$. Total sum $= \dfrac{n}{2}(\text{first}+\text{last}) = \dfrac{29}{2}(1+57) = 29 \times 29 = 841$.

Step 2: Express "sum before $k$" directly in terms of $k$.
The terms strictly before $k$ are $1, 3, \dots, (k-2)$: an AP with common difference $2$, count $= \dfrac{k-1}{2}$, first term $1$, last term $k-2$.
Sum before $= \dfrac{(k-1)/2}{2}\big(1+(k-2)\big) = \dfrac{(k-1)}{4}(k-1) = \dfrac{(k-1)^2}{4}$.

Step 3: Use the equal-halves condition.
Sum before $+$ Sum after $= 841 - k$, and Sum before $=$ Sum after, so:
$2 \times \dfrac{(k-1)^2}{4} = 841 - k \Rightarrow \dfrac{(k-1)^2}{2} = 841-k \Rightarrow (k-1)^2 = 2(841-k)$

Step 4: Solve the equation for $k$.
$k^2 - 2k + 1 = 1682 - 2k$   (the $-2k$ terms cancel from both sides)
$k^2 + 1 = 1682 \Rightarrow k^2 = 1681 \Rightarrow k = 41$ (taking the positive root, since $k$ must be a positive odd number in the sequence).

Final Answer: $k = 41$.