How many terms are there in the series :

$\dfrac{3}{4}, 1, 1\dfrac{1}{4}, ………, 3$ ?

Since, $1 - \dfrac{3}{4} = \dfrac{1}{4}, \dfrac{5}{4} - 1 = \dfrac{1}{4}.$

Hence, the series is an A.P. with common difference = $\dfrac{1}{4}$ and last term = 3.

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

a_n = a + (n - 1)d

$\Rightarrow 3 = \dfrac{3}{4} + (n - 1) \times \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{3 + n - 1}{4} = 3 \\[1em] \Rightarrow n + 2 = 12 \\[1em] \Rightarrow n = 10.$

Hence, no. terms in the series = 10.