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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