Question

Let $\frac{1}{x_1},\frac{1}{x_2},...,\frac{1}{x_n}(x_i\ne 0$ 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 $\sum _{i=1}^n\left(\frac{1}{x_i}\right)$ is equal to

• A. $3$
• B. $\frac{1}{8}$
• C. $\frac{13}{4}$
• D. $\frac{13}{8}$
  

Answer 1

The correct option is C $\frac{13}{4}$

$\frac{1}{4}+20\cdot d=\frac{1}{20}\Rightarrow d=\frac{-1}{100}$

Now, $x_n>50$
$\Rightarrow \frac{1}{x_n}<\frac{1}{50}\Rightarrow \frac{1}{x_n}=\frac{1}{4}-\frac{n-1}{100}<\frac{1}{50}\Rightarrow n>24$
Least value of $n=25$

$\therefore \sum _{i=1}^{25}\left(\frac{1}{x_i}\right)=\frac{25}{2}[2\times \frac{1}{4}-\frac{1}{100}\times 24]=\frac{13}{4}$