A) 3
B) \[\frac{13}{8}\]
C) \[\frac{13}{4}\]
D) \[\frac{1}{8}\]
Correct Answer: C
Solution :
\[\{\frac{1}{{{x}_{n}}},i=1,2,3,...,n\}\]is in AP. \[{{x}_{1}}=4\]and \[{{x}_{21}}=20.\]s By the formula of AP, \[\frac{1}{{{x}_{21}}}=\frac{1}{{{x}_{1}}}+(21-1)\times d\]where d is the common difference. Hence, \[d=\frac{1}{20}\times (\frac{1}{20}-\frac{1}{4})\] \[d=-\frac{1}{100}.\] Also given,\[{{x}_{n}}>50.\] Again using the formula for AP we can write, \[\frac{1}{{{x}_{n}}}=\frac{1}{{{x}_{1}}}+(n-1)\times d\] \[{{x}_{n}}=\frac{{{x}_{1}}}{1+(n-1)\times d\times {{x}_{1}}}\] Therefore, \[\frac{{{x}_{1}}}{1+(n-1)\times d\times {{x}_{1}}}>50\] \[\frac{4}{1+(n-1)\times (-\frac{1}{100}\times 4)}>50\] \[1+(n-1)\times (-\frac{1}{100})\times 4<\frac{4}{50}\] \[-\frac{1}{100}(n-1)<-\frac{23}{100}\] \[n-1>23\] \[n>24\] Therefore, \[n=25.\] \[\sum{_{i=1}^{289}}\frac{1}{{{x}_{i}}}=\frac{25}{2}(2\times \frac{1}{4}+(25-1)\times (-\frac{1}{100})\] \[=\frac{13}{4}\] Option C is correct.
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