Question

Let $\frac{1}{x_1},\frac{1}{x_2},.,\frac{1}{x_n}(x_i\ne 0fori=1,2,..,n)$ be in $A.P.$ such that $x_1=4$ and $x_{21}=20$. If $n$ is the least positive integer fort which $x_n>50$, then $\sum _{i=1}^n\left(\frac{1}{x}\right)$ is equal to:

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

Answer 1

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

We have,

$\frac{1}{x_1},\frac{1}{x_2},\frac{1}{x_3},....,\frac{1}{x_{20}},\frac{1}{20}$

$\frac{1}{20}=\frac{1}{4}+20d$

$d=\frac{-1}{100}$

Now,

$x_n>50$

$\frac{1}{x_n}<\frac{1}{50}$

$\frac{1}{4}-\frac{1}{100}\left(n-1\right)<\frac{1}{30}$

$\frac{26-n}{100}<\frac{1}{50}$

$26-n<2$

$24>n$ $\therefore n=25$

Now,

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

$=\frac{25}{2}\times \frac{26}{100}$

$=\frac{13}{4}$

Hence the option $\left(D\right)$ is the correct answer.