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 $A.P.$ with $x_1 = 4$ and $x_{21} = 20$.Let $d$ be the common difference of

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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 $A.P.$ with $x_1 = 4$ and $x_{21} = 20$.Let $d$ be the common difference of this $A.P.$Then,$\frac{1}{x_{21}} = \frac{1}{x_1} + (21 - 1)d$.$\frac{1}{20} = \frac{1}{4} + 20d \implies 20d = \frac{1}{20} - \frac{1}{4} = \frac{1-5}{20} = -\frac{4}{20} = -\frac{1}{5}$.So,$d = -\frac{1}{100}$.The $n$-th term of the $A.P.$ is $\frac{1}{x_n} = \frac{1}{x_1} + (n-1)d = \frac{1}{4} - \frac{n-1}{100} = \frac{25 - n + 1}{100} = \frac{26 - n}{100}$.Thus,$x_n = \frac{100}{26 - n}$.We are given $x_n > 50$,so $\frac{100}{26 - n} > 50$.Since $x_n > 0$,$26 - n$ must be positive,so $n Dividing by $50$,we get $\frac{2}{26 - n} > 1 \implies 2 > 26 - n \implies n > 24$.The least positive integer $n$ satisfying $n > 24$ is $n = 25$.Now,we calculate the sum $S_n = \sum_{i=1}^n \frac{1}{x_i} = \frac{n}{2} [2a + (n-1)d]$,where $a = \frac{1}{4}$ and $d = -\frac{1}{100}$.$S_{25} = \frac{25}{2} [2(\frac{1}{4}) + (25-1)(-\frac{1}{100})] = \frac{25}{2} [\frac{1}{2} - \frac{24}{100}] = \frac{25}{2} [\frac{50 - 24}{100}] = \frac{25}{2} [\frac{26}{100}] = \frac{25}{2} 	imes \frac{13}{50} = \frac{13}{4}$.

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 $\sum_{i=1}^n \frac{1}{x_i}$ is equal to:

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