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The sum of the first \(n\) terms of an arithmetic sequence is given by \(S_n = \frac{n}{2}(2a + (n-1)d)\). If the first term \(a = 5\) the common difference \(d = 3\) and \(S_n = 155\) find \(n\).

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We substitute \(a = 5\) \(d = 3\) and \(S_n = 155\) into the formula:

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\(155 = \frac{n}{2}(2 \times 5 + (n-1) \times 3)\).

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Simplify inside the parentheses: \(155 = \frac{n}{2}(10 + 3n - 3) = \frac{n}{2}(3n + 7)\).

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Multiply both sides by 2: \(310 = n(3n + 7)\).

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Expand and rearrange: \(3n^2 + 7n - 310 = 0\).

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Use the quadratic formula: \(n = \frac{-7 \pm \sqrt{7^2 - 4 \times 3 \times (-310)}}{2 \times 3}\).

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Calculate the discriminant: \(\Delta = 49 + 3720 = 3769\).

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\(\sqrt{3769} = 61\) so \(n = \frac{-7 \pm 61}{6}\).

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Possible solutions: \(n = \frac{54}{6} = 9\) and \(n = \frac{-68}{6}\) (not valid as \(n\) must be positive).

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#### \(n = 9\)

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