Let S_k = frac{1 + 2 + 3 + .... + k}{k}. If S | vedclass

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(C) The sum of the first $k$ natural numbers is given by $\frac{k(k+1)}{2}$.Thus,$S_k = \frac{k(k+1)}{2k} = \frac{k+1}{2}$.We are given the equation $

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$303$

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(C) The sum of the first $k$ natural numbers is given by $\frac{k(k+1)}{2}$.Thus,$S_k = \frac{k(k+1)}{2k} = \frac{k+1}{2}$.We are given the equation $\sum_{k=1}^{10} S_k^2 = \frac{5}{12}A$.Substituting $S_k$,we get $\sum_{k=1}^{10} \left(\frac{k+1}{2}ight)^2 = \frac{5}{12}A$.This expands to $\frac{1}{4} \sum_{k=1}^{10} (k+1)^2 = \frac{5}{12}A$.Let $n = k+1$. When $k=1, n=2$; when $k=10, n=11$. So,$\frac{1}{4} \sum_{n=2}^{11} n^2 = \frac{5}{12}A$.We know $\sum_{n=1}^{11} n^2 = \frac{11(11+1)(2 	imes 11 + 1)}{6} = \frac{11 	imes 12 	imes 23}{6} = 11 	imes 2 	imes 23 = 506$.Therefore,$\sum_{n=2}^{11} n^2 = 506 - 1^2 = 505$.Substituting this back: $\frac{1}{4} 	imes 505 = \frac{5}{12}A$.$A = 505 	imes \frac{12}{4 	imes 5} = 505 	imes \frac{3}{5} = 101 	imes 3 = 303$.

Let $S_k = \frac{1 + 2 + 3 + .... + k}{k}$. If $S_1^2 + S_2^2 + ....... + S_{10}^2 = \frac{5}{12}A$,then $A$ is equal to

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