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?

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For consecutive odd numbers, remember: sum of first \(n\) terms = \(n^2\). This simplifies balance-sum problems significantly.
Updated On: Dec 5, 2025
  • \(37\)
  • \(41\)
  • \(39\)
  • \(43\)
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The Correct Option is B

Solution and Explanation

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: \begin{align*} 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. \end{align*} Factoring: \[ (m - 21)(m + 20) = 0. \] So, \(m = 21\). Then, \[ k = 2m - 1 = 41. \]
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