The numbers (sqrt 2 + 1),;1,;(sqrt 2 - 1) wil | vedclass

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Question

(b) The numbers $(\sqrt 2 + 1),\;1,\;(\sqrt 2 - 1)$ will be in $G.P.$

$	herefore $ ${(1)^2} = (\sqrt 2 + 1)(\sqrt 2 - 1) = {(\sqrt 2 )^2} - {(1)^2

Vedclass · Accepted Answer

$G.P.$

Vedclass · Answer

(b) The numbers $(\sqrt 2 + 1),\;1,\;(\sqrt 2 - 1)$ will be in $G.P.$

$	herefore $ ${(1)^2} = (\sqrt 2 + 1)(\sqrt 2 - 1) = {(\sqrt 2 )^2} - {(1)^2} = 2 - 1 = 1$.

The numbers $(\sqrt 2 + 1),\;1,\;(\sqrt 2 - 1)$ will be in

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