Question

Let $\frac{1}{x_1},\frac{1}{x^2},....,\frac{1}{x_n}(x_1\ne 0$ 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. $\frac{1}{8}$
• 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, $\frac{1}{4}+20d=\frac{1}{20}$
$20d=\frac{1}{20}-\frac{1}{4}=\frac{1-5}{20}$
$\Rightarrow 20d=\frac{-4}{20}$
$\Rightarrow d=\frac{-4}{20\times 2}$
$\Rightarrow d=\frac{-1}{100}$
Now, $\frac{1}{x_n}<\frac{1}{50}$
$\frac{1}{4}-\frac{n-1}{100}<\frac{1}{50}$
$\Rightarrow n>24$
$n=25$
Therefore, $\sum _{i=1}^{25}\left(\frac{1}{x_i}\right)=\frac{25}{2}\left(2\times \frac{1}{4}-\frac{1}{100}\times 24\right)$
$=\frac{13}{4}$