JEE Mains · Maths · STD 11 - 8. sequence and series
Let \({S_k} = \frac{{1 + 2 + 3 + .... + k}}{k}\). If \(S_1^2 + S_2^2 + ....... + S_{10}^2 = \frac{5}{{12}}A\), then \(A\) equal to
- A \(283\)
- B \(301\)
- C \(303\)
- D \(156\)
Answer & Solution
Correct Answer
(C) \(303\)
Step-by-step Solution
Detailed explanation
\({S_k} = \frac{{k + 1}}{2}\) \({\sum\limits_{k = 1}^{10} {\left( {\frac{{k + 1}}{2}} \right)} ^2} = \frac{5}{{12}}A\) \({2^2} + {3^2} + {......11^2} = \frac{{5A}}{3}\) \(\frac{{11 \times 12 \times 23}}{6} - 1 = \frac{{5A}}{3}\) \(505 \times \frac{3}{5} = A\) \(A = 303\)
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