Find the 50^th term of the sequence :

$\dfrac{1}{n}, \dfrac{n + 1}{n}, \dfrac{2n + 1}{n}, ……$

Since, $\dfrac{n + 1}{n} - \dfrac{1}{n} = \dfrac{n}{n} = 1 \text{ and } \dfrac{2n + 1}{n} - \dfrac{n + 1}{n} = \dfrac{n}{n}$ = 1.

Hence, the series is an A.P. with common difference = 1

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

⇒ a_n = a + (n - 1)d, where a is the first term.

So, 50^th term is,

$\Rightarrow a_{50} = \dfrac{1}{n} + (50 - 1) \times 1 \\[1em] = \dfrac{1}{n} + 49.$

Hence, 50^th term of the sequence = $\dfrac{1}{n} + 49.$