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Let $\frac{1}{x_{1}},\frac{1}{x_{2}},........,\frac{1}{x_{n}}$

$(x_{i}\neq 0 for, i= 1,2....,n)$

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 $\sum_{i=1}^{n}\left ( \frac{1}{x_{i}} \right )$ is equal to :

Option 1)
1/8

Option 2)
3

Option 3)
13/8

Option 4)
13/4

Answers (2)

best_answer

As we have learned

General term of an A.P. -

$T_{n}= a+\left ( n-1 \right )d$

- wherein

$a\rightarrow$ First term

$n\rightarrow$ number of term

$d\rightarrow$ common difference

$1/x_{1},1/x_{2},1/x_{3},.....,1/x_{n}$

are in A.P

so first term $= 1/4$

and $1/x_{21}= 1/20$

$T_{21}= 1/20= 1/4+20d$

$-1/5 = 20d \Rightarrow d= -1/100$

thus

$1/x_{n}=1/4+(x-1)(-1/100)< 1/50$

$x-1> 23\Rightarrow n> 24$

thus n=25

$S_{25}= 1/4+1/4-1/100+1/4-2/100+.......25terms$

$= 25/4-\left ( \frac{1+2+3+.......+24}{100} \right )= 13/4$

Option 1)

1/8

Option 2)

3

Option 3)

13/8

Option 4)

13/4

Posted by

Himanshu

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