Let S_1, S_2, dots, S_{101} be the consecutiv | vedclass

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(A) Let the common difference of the $A.P.$ be $d$. Then $S_{n+1} - S_n = d$ for all $n$. The given sum is $\sum_{n=1}^{100} \frac{1}{S_n S_{n+1}} = \

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$10$

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(A) Let the common difference of the $A.P.$ be $d$. Then $S_{n+1} - S_n = d$ for all $n$. The given sum is $\sum_{n=1}^{100} \frac{1}{S_n S_{n+1}} = \frac{1}{d} \sum_{n=1}^{100} (\frac{1}{S_n} - \frac{1}{S_{n+1}}) = \frac{1}{d} (\frac{1}{S_1} - \frac{1}{S_{101}}) = \frac{1}{d} (\frac{S_{101} - S_1}{S_1 S_{101}}) = \frac{1}{6}$. Since $S_{101} = S_1 + 100d$, we have $S_{101} - S_1 = 100d$. Substituting this into the equation: $\frac{1}{d} (\frac{100d}{S_1 S_{101}}) = \frac{100}{S_1 S_{101}} = \frac{1}{6}$, so $S_1 S_{101} = 600$. We are given $S_1 + S_{101} = 50$. Let $x = S_1$ and $y = S_{101}$. Then $x + y = 50$ and $xy = 600$. The difference $|x - y| = \sqrt{(x+y)^2 - 4xy} = \sqrt{50^2 - 4(600)} = \sqrt{2500 - 2400} = \sqrt{100} = 10$.

Let $S_1, S_2, \dots, S_{101}$ be the consecutive terms of an $A.P.$ If $\frac{1}{S_1 S_2} + \frac{1}{S_2 S_3} + \dots + \frac{1}{S_{100} S_{101}} = \frac{1}{6}$ and $S_1 + S_{101} = 50$, then $|S_1 - S_{101}|$ is equal to

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