Question

Find the ${12}^{th}$ term of the A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2}$ ..........

Answer 1

Given series is $\sqrt{2},3\sqrt{2},5\sqrt{2},.....$

We know, $t_n=a+(n-1)d$

Thus, $a=\sqrt{2},d=2\sqrt{2}$
Therefore, $t_{12}=\sqrt{2}+(12-1)2\sqrt{2}$
$=\sqrt{2}+\left(11\right)2\sqrt{2}$

$=23\sqrt{2}$

Therefore, ${12}^{th}$ term is $23\sqrt{2}$.