No integer \( n \) satisfies? But problem implies one.

February 22, 2026

$No integer \( n \) satisfies? But problem implies one.$

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Wait: \( n(n+2) = 210 \), try factor pairs: \( 14 \times 15 = 210 \), so \( n + 1 = 15 \), \( n = 14 \)? No: equation is \( n(n + 2) = n^2 + 2n = 210 \).
\( 14 \times 16 = 224 \), \( 13 \times 15 = 195 \), no integer solution? But 13.5 gives ~210.
Actually, solve exactly: \( n = \frac{-2 + \sqrt{4 + 840}}{2} = \frac{-2 + \sqrt{844}}{2} = \frac{-2 + 2\sqrt{211}}{2} = -1 + \sqrt{211} \approx -1 + 14.525 = 13.525 \).
Re-check: \( S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(6 + (n-1)2) = \frac{n}{2}(2n + 4) = n(n + 2) \). Set to 210: \( n^2 + 2n - 210 = 0 \).
Discriminant: \( 4 + 840 = 844 \), not a perfect square. But 13.525 not integer.
Perhaps typo in setup? But assume exact: no integer solution.

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