⚡ AZUCATION SHORTCUT:
Total terms $n = 29 \implies$ Total sum = $29^2 = 841$.
Let sum before $k$ be $S_1$. Then $2S_1 + k = 841$.
Using $S_1 = m^2$ and $k = 2m+1$:
$2m^2 + 2m + 1 = 841 \implies m(m+1) = 420 = 20 \times 21$.
$m=20 \implies k = 2(20)+1 = \mathbf{41}$.

Step 1: Analyze the Series
The set $\{1, 3, 5, \dots, 57\}$ is an AP with first term $a=1$ and common difference $d=2$.
Number of terms ($n$): $57 = 1 + (n-1)2 \implies n=29$.
Sum of first $n$ odd numbers = $n^2$. Total Sum $= 29^2 = 841$.

Step 2: Formulate the Equality
Let $k$ be the $(m+1)^{th}$ term.
Sum of elements less than $k$ (first $m$ terms) $= m^2$.
According to the problem: $Sum_{k}$.
Since $Sum_{k} = Total Sum$, we have:
$m^2 + (2m+1) + m^2 = 841 \implies 2m^2 + 2m + 1 = 841$.

Step 3: Solve for $m$ and $k$
$2m^2 + 2m - 840 = 0 \implies m^2 + m - 420 = 0$.
Factoring: $(m+21)(m-20) = 0$. Since $m > 0$, $m = 20$.
The value of $k$ is the $(20+1)^{th}$ term: $k = 2(20) + 1 = \mathbf{41}$.

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