Question

Which of the following are APs? If they form an A.P. find the common difference $d$ and write three more terms.

(v) $3,3+\sqrt{2},3+2\surd 2,3+3\sqrt{2}$.

Answer 1

Arithmetic progression:

The given series is $3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2}$.

The difference between each term of the given series is,

$$\begin{gathered} 3+\sqrt{2}-3=\sqrt{2} \\ \left(3+2\sqrt{2}\right)-\left(3+\sqrt{2}\right)=\sqrt{2} \\ \left(3+3\sqrt{2}\right)-\left(3+2\sqrt{2}\right)=\sqrt{2} \end{gathered}$$

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:

$$\begin{gathered} \left(3+3\sqrt{2}\right)+\sqrt{2}=3+4\sqrt{2} \\ \left(3+4\sqrt{2}\right)+\sqrt{2}=3+5\sqrt{2} \\ \left(3+5\sqrt{2}\right)+\sqrt{2}=3+6\sqrt{2} \end{gathered}$$

Hence, the given sequence is an A.P. and the common difference is $d=\sqrt{2}$ and the next three terms are $3+4\sqrt{2},3+5\sqrt{2},3+6\sqrt{2}$.