Let frac{1}{x_1}, frac{1}{x_2}, frac{1}{x_3}, | vedclass

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(C) Given that $\frac{1}{x_1}, \frac{1}{x_2}, \dots, \frac{1}{x_n}$ are in $A.P.$Let $a = \frac{1}{x_1} = \frac{1}{4}$ and $d$ be the common differenc

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$\frac{13}{4}$

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(C) Given that $\frac{1}{x_1}, \frac{1}{x_2}, \dots, \frac{1}{x_n}$ are in $A.P.$Let $a = \frac{1}{x_1} = \frac{1}{4}$ and $d$ be the common difference.We have $\frac{1}{x_{21}} = \frac{1}{20}$.Using the $A.P.$ formula $\frac{1}{x_{21}} = a + 20d$,we get $\frac{1}{20} = \frac{1}{4} + 20d$.$20d = \frac{1}{20} - \frac{1}{4} = \frac{1-5}{20} = -\frac{4}{20} = -\frac{1}{5}$.So,$d = -\frac{1}{100}$.The $n^{th}$ term of the $A.P.$ is $\frac{1}{x_n} = a + (n-1)d = \frac{1}{4} - \frac{n-1}{100} = \frac{25 - n + 1}{100} = \frac{26 - n}{100}$.Thus,$x_n = \frac{100}{26 - n}$.Given $x_n > 50$,we have $\frac{100}{26 - n} > 50$.Since $x_n > 50$,$26 - n$ must be positive and $26 - n  24$.The least positive integer $n$ is $25$.We need to find $\sum_{i=1}^{25} \frac{1}{x_i} = \frac{25}{2} \left[ 2a + (25-1)d ight]$.$= \frac{25}{2} \left[ 2 \left( \frac{1}{4} ight) + 24 \left( -\frac{1}{100} ight) ight] = \frac{25}{2} \left[ \frac{1}{2} - \frac{6}{25} ight] = \frac{25}{2} \left[ \frac{25 - 12}{50} ight] = \frac{25}{2} 	imes \frac{13}{50} = \frac{13}{4}$.

Let $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}, \dots, \frac{1}{x_n}$ ($x_i \neq 0$ for $i = 1, 2, \dots, n$) be in $A.P.$ such that $x_1 = 4$ and $x_{21} = 20$. If $n$ is the least positive integer for which $x_n > 50$,then $\sum_{i=1}^n \left( \frac{1}{x_i} \right)$ is equal to:

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