Question

Is the given sequence $3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2}$ form an APs? If it forms an AP, find the common difference $d$ and write the next three terms.

Answer 1

$t_1=3$

$t_2=3+\sqrt{2}$

$t_3=3+3\sqrt{2}$

$t_4=3+3\sqrt{2}$

$t_2-t_1=\sqrt{2}$

$t_3-t_2=3+2\sqrt{2}-(3+\sqrt{2})=\sqrt{2}$

$t_4-t_3=3+3\sqrt{2}-(3+2\sqrt{2})=\sqrt{2}$

Since, the difference between two consecutive terms is constant.

$\therefore$ sequences is in A.P.

And, the next three terms are

$t_5=3+3\sqrt{2}+\sqrt{2}=3+4\sqrt{2}$

$t_6=3+4\sqrt{2}+\sqrt{2}=3+5\sqrt{2}$

$t_7=3+5\sqrt{2}+\sqrt{2}=3+6\sqrt{2}$