Letn frac1x_1,frac1x_2, ... ,frac1x_n( x_i ≠ 0 . n for . i = 1,2, ... ,n ) be in A.P. such thatn x_1 = 4 and x_21 = 20. If n is the least positiven integer for which x_n > 50, thenn sum _i = 1^nmspace2mu( frac1x_i ) isn equal to

n n (d) : x_1=4 and x_21=20 n => (1 / x_1)=(1 / 4) and frac1x_21=(1 / 20)n nn nNow, a_21=a+20 d=(1 / 20)(Because a_21=(1 / 20))nn => (1 / 4)+20 d=(1 / 20) => d=-(1 / 100)nn nNow, x_n>50 (Given) => (1 / x_n)24nn nNow, sum _i=1^n((1 / x_i))=(n / 2)[2 a+(n-1) d]nn => sum _i=1^25((1 / x_i))=(25 / 2)[2 x (1 / 4)-(1 / 100) x 24]=(13 / 4)n$

Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}... | Sequence & Series (Mathematics)

\n\begin{aligned}\n& \text { (d) }: x_1=4 \text { and } x_{21}=20 \\\n& \Rightarrow \frac{1}{x_1}=\frac{1}{4} \text { and } \frac{1}{x_{21}}=\frac{1}{20}\n\end{aligned}\n

\n \nNow,

a_{21}=a+20 d=\frac{1}{20}\left(\because a_{21}=\frac{1}{20}\right)

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

\n \nNow,

x_n>50

(Given)

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

\n\Rightarrow \frac{1}{4}-\frac{(n-1)}{100}<\frac{1}{50} \Rightarrow n>24\n

\n \nNow,

\sum_{i=1}^n\left(\frac{1}{x_i}\right)=\frac{n}{2}[2 a+(n-1) d]

\n\Rightarrow \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}\n

Explanation

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