MCQ
The numbers $(\sqrt 2 + 1),\;1,\;(\sqrt 2 - 1)$ will be in
- A$A.P.$
- ✓$G.P.$
- C$H.P.$
- DNone of these
$\therefore $ ${(1)^2} = (\sqrt 2 + 1)(\sqrt 2 - 1) = {(\sqrt 2 )^2} - {(1)^2} = 2 - 1 = 1$.
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