Find the 100^th term of the sequence :

$\sqrt{3}, 2\sqrt{3}, 3\sqrt{3},$ ……..

Since, $2\sqrt{3} - \sqrt{3} = \sqrt{3}, 3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$.

Hence, the series is an A.P. with common difference = $\sqrt{3}$.

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, 100^th term is,

$\Rightarrow a_{100} = \sqrt{3} + (100 - 1) \times \sqrt{3} \\[1em] = \sqrt{3} + 99 \times \sqrt{3} \\[1em] = 100\sqrt{3}.$

Hence, 100^th term of the sequence = $100\sqrt{3}$.