Question:
Find the $\mathrm{n}^{\text {th }}$ term of the AP $1, \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \ldots$
Solution:
To Find: $\mathrm{n}^{\text {th }}$ term of the AP
Given: The series is $1, \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \ldots$
$a_{1}=1, a_{2}=\frac{5}{6}$ and $d=\frac{5}{6}-1=\frac{-1}{6}$
(Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given AP)
Formula Used: $a_{n}=a+(n-1) d$
$a_{n}=a_{1}+(n-1)\left(\frac{-1}{6}\right)=1-\left(\frac{n-1}{6}\right)=\frac{6-n+1}{6}=\left(\frac{7-n}{6}\right)$
So the $\mathrm{n}^{\text {th }}$ term of AP is equal to $\frac{7-n}{(6)}$