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

Answer 1

Answer: (C)

$\frac{13}{4}$

$\frac{1}{x_1}=\frac{1}{4};\frac{1}{x_{21}}=\frac{1}{20}$

$\Rightarrow T_1=\frac{1}{4},T_{21}=\frac{1}{20}$

We know that, $T_n=a+(n-1)d$

where $a\to \text{first term,}d\to \text{common difference}$

$\therefore \frac{1}{20}=\frac{1}{4}+(21-1)d$

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

$T_n=\frac{1}{x_n}$ (General term)

$\frac{1}{x_n}=\frac{1}{4}+(n-1)(-\frac{1}{100})$

$\Rightarrow \frac{1}{x_n}=\frac{1}{4}-\frac{n}{100}+\frac{1}{100}$

$\Rightarrow \frac{1}{x_n}=\frac{25-n+1}{100}$

$\Rightarrow x_n=\frac{100}{26-n}$

Given, $x_n>50$

$\therefore \frac{100}{26-n}>50$

$\Rightarrow 100>50(26-n)$

$\Rightarrow n>24$

$\therefore n=\{25,26,27,28,.....\}$

$n=25\text{[least possible integer]}$

$\therefore S_{25}=\frac{25}{2}[2\left(\frac{1}{4}\right)+(25-1\left)(-\frac{1}{100})\right]$ $[S_n=\frac{n}{2}[2a+(n-1)d\left]\right]$

$\Rightarrow S_{25}=\frac{25}{2}[\frac{1}{2}-\frac{24}{100}]=\frac{13}{4}$

Hence, C is the right answer