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

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Question

$\because$ $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},....,\frac{1}{{{x_n}}}$ are in $A.P.$

${x_1} = 4\,\,\,\,\,{x_{21}} = 20

Vedclass · Accepted Answer

$\frac {13}{4}$

Vedclass · Answer

$\because$ $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},....,\frac{1}{{{x_n}}}$ are in $A.P.$ ${x_1} = 4\,\,\,\,\,{x_{21}} = 20$ Let $'d'$ be the common difference of this $A.P.$ $ herefore $ its ${21^{st}}$ term $ = \frac{1}{{{x_{21}}}} = \frac{1}{{{x_1}}} + \left[ {\left( {21 - 1} ight) imes d} ight]$ $ \Rightarrow d = \frac{1}{{20}} imes \left( {\frac{1}{{20}} - \frac{1}{4}} ight) \Rightarrow d = - \frac{1}{{100}}$ Alson ${x_n} > 50$ (given). $ herefore \frac{1}{{{x_n}}} = \frac{1}{{{x_1}}} + \left[ {\left( {n - 1} ight) imes d} ight]$ $ \Rightarrow {x_n} = \frac{{{x_1}}}{{1 + \left( {n - 1} ight) imes d imes {x_1}}}$ $ herefore {x_n} = \frac{{{x_1}}}{{1 + \left( {n - 1} ight) imes d imes {x_1}}} > 50$ $ \Rightarrow \frac{4}{{1 + \left( {n - 1} ight) imes \left( { - \frac{1}{{100}}} ight) imes 4}} > 50$ $ \Rightarrow 1 + \left( {n - 1} ight) imes \left( { - \frac{1}{{100}}} ight) imes 4 < \frac{4}{{50}}$ $ \Rightarrow \left( { - \frac{1}{{100}}} ight)\left( {n - 1} ight) < - \frac{{23}}{{100}}$ $ \Rightarrow n - 1 > 23\,\,\,\,\,\, \Rightarrow n > 24$ Therefore,$n=25$. $ \Rightarrow \sum\limits_{i = 1}^{25} {\frac{1}{x}} = \frac{{25}}{2}\left[ {\left( {2 imes \frac{1}{4}} ight) + \left( {25 - 1} ight) imes \left( { - \frac{1}{{100}}} ight)} ight] = \frac{{13}}{4}$

Let $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},.....,$ $({x_i} \ne \,0\,for\,\,i\, = 1,2,....,n)$ 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\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}}} \right)} $ is equal to.

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