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JEE Mains · Maths · STD 11 - 8. sequence and series

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.

  1. A \(3\)
  2. B \(\frac {13}{8}\)
  3. C \(\frac {13}{4}\)
  4. D \(\frac {1}{8}\)
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Answer & Solution

Correct Answer

(C) \(\frac {13}{4}\)

Step-by-step Solution

Detailed explanation

\(\because\) \(\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},....,\frac{1}{{{x_n}}}\) are in \(A.P.\) \({x_1} = 4\,\,\,\,\,{x_{21}} = 20\) Let \('d'\) be the common difference of this \(A.P.\) \(\therefore \) its \({21^{st}}\) term…
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