Question

In the set of consecutive odd numbers $\{1, 3, 5, \ldots, 57\}$, there is a number $k$ such that the sum of all the elements less than $k$ is equal to the sum of all the elements greater than $k$. Then, $k$ equals?

• A. \(37\)
• B. \(41\)
• C. \(39\)
• D. \(43\)

Hint:

For consecutive odd numbers, remember: sum of first \(n\) terms = \(n^2\). This simplifies balance-sum problems significantly.

Answer 1

The Correct Option is B

Approach: Use the gem fact that the sum of the first \(n\) odd numbers is \(n^2\). If \(k\) is the \(m\)-th odd number, everything before it sums to \((m-1)^2\); set that equal to everything after.

Step 1 (count and total): Odd numbers \(1,3,\dots,57\): there are \(\dfrac{57-1}{2}+1 = 29\) terms, total \(= 29^2 = 841\).

Step 2 (set up the balance): Let \(k\) be the \(m\)-th term, so \(k = 2m - 1\). Sum before \(k\) is the first \((m-1)\) odds \(= (m-1)^2\). Sum after \(k\) \(= 841 - (m-1)^2 - k\). Setting before \(=\) after: \[ (m-1)^2 = 841 - (m-1)^2 - (2m-1). \]

Step 3 (solve): \[ 2(m-1)^2 + (2m-1) = 841 \Rightarrow 2m^2 - 2m + 1 = 841 \Rightarrow m^2 - m - 420 = 0. \] Factor: \((m-21)(m+20) = 0 \Rightarrow m = 21.\)

Step 4: \(k = 2(21) - 1 = 41.\)

Check: Sum below \(41\): \(1+3+\dots+39 = 20^2 = 400\); sum above \(41\): \(43+\dots+57 = 841 - 400 - 41 = 400\). Balanced.

Answer: \(\boxed{41}\)

Answer 2

Step 1: Identify the sequence properties. The sequence is: \[ 1, 3, 5, \dots, 57. \] First term \(a = 1\), last term \(l = 57\). Number of terms: \[ n = \frac{57 - 1}{2} + 1 = 29. \] Sum of first \(n\) odd numbers: \[ 29^2 = 841. \] 
Step 2: Set up the equation. Let \(k\) be the \(m\)-th term. Sum of terms before \(k\): \((m-1)^2\). Sum of terms after \(k\): also \((m-1)^2\). Total sum: \[ 841 = 2(m-1)^2 + (2m - 1). \]
Simplifying: \(2(m-1)^2 + (2m - 1) = 841 \)
\(2m^2 - 4m + 2 + 2m - 1= 841 \)
\(2m^2 - 2m + 1 = 841 \)
\(2m^2 - 2m - 840 = 0 \)
\(m^2 - m - 420= 0.\)
 Factoring: \[ (m - 21)(m + 20) = 0. \] So, \(m = 21\). Then, \[ k = 2m - 1 = 41. \]