Which term of A.P. 1, $1\dfrac{1}{6}, 1\dfrac{1}{3}$,….,is 3?

Series :

$\Rightarrow 1, 1\dfrac{1}{6}, 1\dfrac{1}{3}, ……$

$\Rightarrow 1, \dfrac{7}{6}, \dfrac{4}{3}, ……$

In the A.P. 1, $1\dfrac{1}{6}, 1\dfrac{1}{3}$,…., a = 1 and d = $\dfrac{7}{6} - 1 = \dfrac{1}{6}$

Let n^th term be 3.

∴ a_n = 3

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

$\Rightarrow 3 = 1 + (n - 1)\dfrac{1}{6} \\[1em] \Rightarrow 3 = 1 + (n - 1)\dfrac{1}{6} \\[1em] \Rightarrow 3 - 1 = (n - 1)\dfrac{1}{6} \\[1em] \Rightarrow 2 = (n - 1)\dfrac{1}{6} \\[1em] \Rightarrow 2 \times 6 = (n - 1) \\[1em] \Rightarrow 12 = (n - 1) \\[1em] \Rightarrow n = 12 + 1 \\[1em] \Rightarrow n = 13.$

Hence, 13th term of A.P. is 3.