Question

Let $\frac{1}{x_1} , \frac{1}{x^2} ,...., \frac{1}{x_n} (x_1 \neq 0$ 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 $\displaystyle\sum^n_{i = 1} \left( \frac{1}{x_i}\right)$ is equal to

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

Answer 1

Answer: (D)

Given:
$AP: \frac{1}{x_{1}}, \frac{1}{x_{2}}, \dots \frac{1}{x_{n}}$
And $x_{1}=4, x_{21}=20$
So, $\frac{1}{4}+20d=\frac{1}{20}$
$20 d=\frac{1}{20}-\frac{1}{4}=\frac{1-5}{20} $
$\Rightarrow 20 d=\frac{-4}{20}$
$\Rightarrow d=\frac{-4}{20 \times 2}$
$ \Rightarrow d=\frac{-1}{100}$
Now, $\frac{1}{x_{n}}<\,\frac{1}{50} $
$\frac{1}{4}-\frac{n-1}{100}<\,\frac{1}{50} $
$\Rightarrow n >\,24 $
$n=25 $
Therefore, $\displaystyle\sum_{i=1}^{25}\left(\frac{1}{x_{i}}\right)=\frac{25}{2}\left(2 \times \frac{1}{4}-\frac{1}{100} \times 24\right)$
$=\frac{13}{4}$