Question

Which of the following are APs?

If they form an A.P. find the common difference $d$ and write three more terms.

$\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32}\dots$

Answer 1

Solution.

Given series is $\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32}\dots$.

The difference between each terms of the given series is,

$$\begin{gathered} \sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2} \\ =\sqrt{2} \\ \sqrt{18}-\sqrt{8}=3\sqrt{2}-2\sqrt{2} \\ =\sqrt{2} \\ 4\sqrt{2}-3\sqrt{2}=\sqrt{2} \end{gathered}$$

Since, the common difference is same every time.

Therefore, $d=\sqrt{2}$ and the given series forms a A.P.

So, the next three terms are;

$$\begin{gathered} \sqrt{32}+\sqrt{2}=4\sqrt{2}+\sqrt{2} \\ =\sqrt{50} \\ \sqrt{50}+\sqrt{2}=5\sqrt{2}+\sqrt{2} \\ =\sqrt{72} \\ \sqrt{72}+\sqrt{2}=6\sqrt{2}+\sqrt{2} \\ =\sqrt{98} \end{gathered}$$

Hence, the given series is an A.P. and the common difference is $d=\sqrt{2}$ and the next three terms are $\sqrt{50},\sqrt{72},\sqrt{98}$.