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?
Show Hint
- \(37\)
- \(41\)
- \(39\)
- \(43\)
The Correct Option is B
Approach Solution - 1
Step 1: Analyze the given sequence. The sequence consists of consecutive odd numbers: \[ 1, 3, 5, \dots, 57. \] Here, the first term is \(1\) and the last term is \(57\). The total number of terms is \[ n = \frac{57 - 1}{2} + 1 = 29. \] The sum of the first 29 odd numbers is \[ 29^2 = 841. \] Step 2: Form the required equation. Let the required middle term be the \(m\)-th term of the sequence. The sum of the terms before this term is \[ (m - 1)^2, \] and the sum of the terms after it is also \[ (m - 1)^2. \] Hence, the total sum of the sequence can be written as \[ 841 = 2(m - 1)^2 + (2m - 1). \] Simplifying, \[ 2(m^2 - 2m + 1) + 2m - 1 = 841, \] \[ 2m^2 - 4m + 2 + 2m - 1 = 841, \] \[ 2m^2 - 2m + 1 = 841, \] \[ 2m^2 - 2m - 840 = 0. \] Dividing throughout by 2, \[ m^2 - m - 420 = 0. \] Factoring, \[ (m - 21)(m + 20) = 0. \] Thus, \(m = 21\). Step 3: Find the required term. The \(m\)-th term of the sequence is \[ k = 2m - 1 = 2(21) - 1 = 41. \]
Approach Solution -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, \[ k = 2m - 1 = 41. \]
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