Which term of the AP

In the given $\mathrm{AP}$, first term $=\frac{5}{6}$ and common difference, $\mathrm{d}=\left(1-\frac{5}{6}=\frac{1}{6}\right)$.

Let its nth term be 3.

Now, $T_{n}=3$

$\Rightarrow a+(n-1) d=3$

$\Rightarrow \frac{5}{6}+(n-1) \times \frac{1}{6}=3$

$\Rightarrow \frac{2}{3}+\frac{n}{6}=3$

$\Rightarrow \frac{n}{6}=\frac{7}{3}$

$\Rightarrow n=14$

Hence, the 14th term of the given AP is 3.