Question:
Find the $20^{\text {th }}$ term of the AP $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots .$
Solution:
To Find: $20^{\text {th }}$ term of the AP
Given: The series is $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots$
$a_{1}=\sqrt{2}, a_{2}=3 \sqrt{2}$ and $d=3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}$
(Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $A P$ )
Formula Used: $a_{n}=a+(n-1) d$
$a_{20}=a_{1}+(20-1)(2 \sqrt{2})=\sqrt{2}+38 \sqrt{2}=39 \sqrt{2}$
So $20^{\text {rd }}$ term of AP is equal to $39 \sqrt{2}$.