JEE Main On 16 April 2018 Question 3
Question: Let $ \frac{1}{x _1},\frac{1}{x _2},….,\frac{1}{x _{n}}(x _{i}\ne 0for,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}{( \frac{1}{x _{i}} )} $ is equal to. [JEE Main 16-4-2018]
Options:
A) 3
B) $ \frac{13}{8} $
C) $ \frac{13}{4} $
D) $ \frac{1}{8} $
Show Answer
Answer:
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.
