To calculate \( S_{58} \) for the arithmetic sequence defined by \( a_n = \frac{5}{6} n + \frac{1}{3} \), we first need to determine the first term (\( a_1 \)), the last term (\( a_{58} \)), and the number of terms (which is 58 in this case).
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Calculate the first term, \( a_1 \): \[ a_1 = a(1) = \frac{5}{6} \cdot 1 + \frac{1}{3} = \frac{5}{6} + \frac{2}{6} = \frac{7}{6} \]
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Calculate the last term, \( a_{58} \): \[ a_{58} = a(58) = \frac{5}{6} \cdot 58 + \frac{1}{3} = \frac{290}{6} + \frac{2}{6} = \frac{292}{6} = \frac{146}{3} \]
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Calculate the number of terms, \( n \): The number of terms \( n \) in this case is 58.
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Calculate the sum \( S_n \) for an arithmetic sequence: \[ S_n = \frac{n}{2} (a_1 + a_n) \] Substituting in our values: \[ S_{58} = \frac{58}{2} \left(\frac{7}{6} + \frac{146}{3}\right) \] First, we need a common denominator to add \( \frac{7}{6} \) and \( \frac{146}{3} \). The common denominator is 6.
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Convert \( \frac{146}{3} \) to sixths: \[ \frac{146}{3} = \frac{146 \cdot 2}{3 \cdot 2} = \frac{292}{6} \]
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Now add \( a_1 \) and \( a_{58} \): \[ \frac{7}{6} + \frac{292}{6} = \frac{7 + 292}{6} = \frac{299}{6} \]
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Finally, calculate \( S_{58} \): \[ S_{58} = \frac{58}{2} \cdot \frac{299}{6} = 29 \cdot \frac{299}{6} = \frac{29 \cdot 299}{6} = \frac{8671}{6} \]
Thus, the result is:
\[ \boxed{\frac{8671}{6}} \]