Question

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

Answer 1

Given A.P

$1,\frac{21}{11},\frac{41}{11},...\dots \dots .;x=\frac{171}{11}$

first term of this A.P is $a_1=1$

second term of this A.P is $a_2=\frac{21}{11}$

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

common difference of A.P

$d=a_2-a_1=\left(\frac{21}{11}\right)-\left(1\right)=\frac{10}{11}$

the nth term of an A.P is given by

$a_n=a_1+(n-1)d$

put $a_n=\frac{171}{11};a_1=1;d=\frac{10}{11}$ in above equation we get

$⟹\frac{171}{11}=1+(n-1)\left(\frac{10}{11}\right)$

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

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

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

$⟹n=\frac{170\times 11}{10\times 11}$

$⟹n=17$