Insert 5 A.Ms between (\sqrt{2}) and (\dfrac{15}{\sqrt{2}}).

Total terms = 7: (a_1 = \sqrt{2}), (a_7 = \dfrac{15}{\sqrt{2}})

[ \dfrac{15}{\sqrt{2}} = \sqrt{2} + 6d ] [ 6d = \dfrac{15}{\sqrt{2}} - \sqrt{2} = \dfrac{15 - 2}{\sqrt{2}} = \dfrac{13}{\sqrt{2}} ] [ d = \dfrac{13}{6\sqrt{2}} = \dfrac{13\sqrt{2}}{12} ]

The five A.Ms are: [ \begin{align*} \sqrt{2}+d &= \dfrac{25\sqrt{2}}{12} \ \sqrt{2}+2d &= \dfrac{19\sqrt{2}}{6} \ \sqrt{2}+3d &= \dfrac{17\sqrt{2}}{4} \ \sqrt{2}+4d &= \dfrac{16\sqrt{2}}{3} \ \sqrt{2}+5d &= \dfrac{77\sqrt{2}}{12} \end{align*} ]

Answer: (\dfrac{25\sqrt{2}}{12},\ \dfrac{19\sqrt{2}}{6},\ \dfrac{17\sqrt{2}}{4},\ \dfrac{16\sqrt{2}}{3},\ \dfrac{77\sqrt{2}}{12})