Find $n$ if the given value of $x$ is the nth term of the given A.P.
$ 1, \frac{21}{11}, \frac{31}{11}, \frac{41}{11}, \ldots, x=\frac{171}{11} $

Given:

Given A.P. is \( 1, \frac{21}{11}, \frac{31}{11}, \frac{41}{11}, \ldots \)

$x=\frac{171}{11}$ is the nth term of the A.P. 

To do:  

We have to find the value of $n$.

Solution:

We know that,

nth term of an A.P. $a, a+d, a+2d,.....$ is $a_n=a+(n-1)d$.

In the given A.P.,

$a_1=1, a_2=\frac{21}{11}, a_3=\frac{31}{11}$ and common difference $d=\frac{21}{11}-1=\frac{21-1\times11}{11}=\frac{21-11}{11}=\frac{10}{11}$

This implies,

$x=1+(n-1)(\frac{10}{11})$

$\frac{171}{11}=1+\frac{10}{11}n-\frac{10}{11}$

$\frac{10}{11}n=\frac{171}{11}+\frac{10}{11}-1$

$\frac{10}{11}n=\frac{171+10-1\times11}{11}$

$10n=181-11$

$n=\frac{170}{10}$

$n=17$

The value of $n$ is $17$.  

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Updated on: 10-Oct-2022

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