Sequence And Series Question 239
Question: The numbers $ (\sqrt{2}+1),\ 1,\ (\sqrt{2}-1) $ will be in
[AMU 1983]
Options:
A) A.P.
B) G.P.
C) H.P.
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
The numbers $ (a_1+a_{24})+(a_5+a_{20})+(a_{10}+a_{15})=225 $ will be in G.P.
$ \therefore $ $ {{(1)}^{2}}=(\sqrt{2}+1)(\sqrt{2}-1)={{(\sqrt{2})}^{2}}-{{(1)}^{2}}=2-1=1 $ .
BETA
