\( \sqrt{844} = \sqrt{4 \times 211} = 2\sqrt{211} \).
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- General: \( S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[6 + (n-1)2] = \frac{n}{2}(2n + 4) = n(n + 2) \). Correct.
- So \( n(n+2) = 210 \).
- \( n = \frac{ -2 \pm \sqrt{4 + 840} }{2} = \frac{ -2 \pm \sqrt{844} }{2} \).
- \( \sqrt{211} \approx 14.525 \), so \( n \approx \frac{27.05}{2} = 13.525 \).
- Since \( n \) must be integer, and 13 gives 195, 14 gives 224, no solution.
- But the problem likely has a typo.
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