Question:
Let \(S_k = \frac{1 + 2 + 3 + \dots + k}{k}\). If \(S_1^2 + S_2^2 + \dots + S_{10}^2 = \frac{5}{12}A\), then A is equal to :
Let \(S_k = \frac{1 + 2 + 3 + \dots + k}{k}\). If \(S_1^2 + S_2^2 + \dots + S_{10}^2 = \frac{5}{12}A\), then A is equal to :
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Simplify \(S_k = \frac{k+1}{2}\) and use the formula for the sum of squares.
Updated On: Nov 5, 2026
- \(283\)
- \(156\)
- \(301\)
- \(303\)
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The Correct Option is D
Solution and Explanation
\(S_k = \frac{k(k+1)/2}{k} = \frac{k+1}{2}\), so \(S_k^2 = \frac{(k+1)^2}{4}\).
\(\sum_{k=1}^{10} S_k^2 = \frac{1}{4}\left(2^2 + 3^2 + \dots + 11^2\right) = \frac{1}{4}\left(\frac{11\cdot12\cdot23}{6} - 1\right) = \frac{1}{4}(506 - 1) = \frac{505}{4}\).
Given \(\frac{505}{4} = \frac{5}{12}A \Rightarrow A = \frac{505 \times 12}{4 \times 5} = 303\).
Answer: (4) 303
\(\sum_{k=1}^{10} S_k^2 = \frac{1}{4}\left(2^2 + 3^2 + \dots + 11^2\right) = \frac{1}{4}\left(\frac{11\cdot12\cdot23}{6} - 1\right) = \frac{1}{4}(506 - 1) = \frac{505}{4}\).
Given \(\frac{505}{4} = \frac{5}{12}A \Rightarrow A = \frac{505 \times 12}{4 \times 5} = 303\).
Answer: (4) 303
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