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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

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 (-1) 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 $a=\frac{1}{2}-\frac{1}{\sqrt{2}}.$