Find the 100^th term of the sequence :

$\sqrt{5}, 2\sqrt{5}, 3\sqrt{5}, …….$

In the sequence :

$3\sqrt{5} - 2\sqrt{5} = 2\sqrt{5} - \sqrt{5} = \sqrt{5}$.

Since, the difference between consecutive terms are equal, thus the sequence is an A.P.

First term (a) = $\sqrt{5}$

Common difference (d) = $\sqrt{5}$

We know that,

$\Rightarrow a$n = a + (n - 1)d \\[1em] \Rightarrow a{100} = \sqrt{5} + (100 - 1) \times \sqrt{5} \\[1em] \Rightarrow a{100} = \sqrt{5} + 99\sqrt{5} \\[1em] \Rightarrow a{100} = 100\sqrt{5}.