Question

Is the sequence $\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},......$ from an Arithmetic Progression?Give reason.

Answer 1

$\sqrt{3},\sqrt{6},\sqrt{9},.....$

$\sqrt{6}-\sqrt{3}=\sqrt{3}(\sqrt{2}-1)$

$\sqrt{9}-\sqrt{6}=\sqrt{3}(\sqrt{3}-\sqrt{2})$

Since the difference between the two consecutive terms of the given series is not same, thus the given series is not an arithmetic progression.