Gujarati
8. Sequences and Series
easy

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

A

$A.P.$

B

$G.P.$

C

$H.P.$

D

None of these

Solution

(b) The numbers $(\sqrt 2 + 1),\;1,\;(\sqrt 2 – 1)$ will be in $G.P.$

$\therefore $ ${(1)^2} = (\sqrt 2 + 1)(\sqrt 2 – 1) = {(\sqrt 2 )^2} – {(1)^2} = 2 – 1 = 1$.

Standard 11
Mathematics

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