Which term of the AP
Question:

Which term of the AP $\frac{5}{6}, 1,1 \frac{1}{6}, 1 \frac{1}{3}, \ldots .$ is 3 ?

 

Solution:

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.

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