Sequence And Series Question 429
Question: a, b, c are three distinct real numbers and they are in a GP. if a + b + c = xb, then
Options:
A) $ x\le -1orx\ge 3 $
B) $ x<-1orx>3 $
C) $ x\le -1orx>3 $
D) $ x<-3orx>2 $
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Answer:
Correct Answer: B
Solution:
[b] Let $ b=ar $ and $ c=ar^{2}, $ so that a, b, c are in GP.
$ \therefore a+b+c=xb $
$ \Rightarrow ,a+ar+ar^{2}=x.ar\Rightarrow ,r^{2}+(1-x)r+1=0 $ ? (1) If r is real, then discriminant of $ (1)\ge 0 $
$ \Rightarrow ,{{(1-x)}^{2}}-4.1.1\ge 0\Rightarrow x^{2}-2x-3\ge 0 $
$ \Rightarrow ,(x+1)(x-3)\ge 0\Rightarrow x\le -1 $ or $ x\ge 3. $ Now for $ x=3 $ we get $ r=1, $ which will make $ a=b=c $ Also for $ x=-1, $ we get $ r=-1, $ for which $ a=c, $ thus $ x<-1 $ or $ x>3 $
BETA
