Sequence And Series Question 505
Question: Suppose $ a,,b,,c $ are in A.P. and $ a^{2},b^{2},c^{2} $ are in G.P. If a < b < c and $ a+b+c=\frac{3}{2} $ , then the value of a is
[IIT Screening 2002]
Options:
A) $ \frac{1}{2\sqrt{2}} $
B) $ \frac{1}{2\sqrt{3}} $
C) $ \frac{1}{2}-\frac{1}{\sqrt{3}} $
D) $ \frac{1}{2}-\frac{1}{\sqrt{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ b=a+d,c=a+2d $ , where $ d>0 $ Now $ a^{2},{{(a+d)}^{2}},{{(a+2d)}^{2}} $ are in G.P.
$ \therefore {{(a+d)}^{4}}=a^{2}{{(a+2d)}^{2}} $ or $ {{(a+d)}^{2}}=\pm a(a+2d) $ or $ a^{2}+d^{2}+2ad=\pm (a^{2}+2ad) $ Taking (+) sign, d = 0 (not possible as $ a<b<c) $ Taking (-) sign, $ 2a^{2}+4ad+d^{2}=0 $ , $ [ \because a+b+c=\frac{3}{2},,\therefore a+d=\frac{1}{2} ] $ $ 2a^{2}+4a( \frac{1}{2}-a )+{{( \frac{1}{2}-a )}^{2}}=0 $ or $ 4a^{2}-4a-1=0 $
$ \therefore a=\frac{1}{2}\pm \frac{1}{\sqrt{2}}.Here,d=\frac{1}{2}-a>0. $ So, $ a<\frac{1}{2}. $ Hence $ a=\frac{1}{2}-\frac{1}{\sqrt{2}} $ .
BETA
