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.
- A \(3\)
- B \(\frac {13}{8}\)
- C \(\frac {13}{4}\)
- D \(\frac {1}{8}\)
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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