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 value of a is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$b=a_1$

$a_1-d+a_1+a_1+d=\frac{3}{2}$

$c=a_1+d$

$3a_1=\frac{3}{2}$

$a_1=\frac{1}{2}$

$\frac{b^2}{a^2}=\frac{c^2}{b^2}$

$\Rightarrow b^4=a^2{c}^2$

$\Rightarrow {\left(\frac{1}{2}\right)}^4={\left[(\frac{1}{2}+d)(\frac{1}{2}-d)\right]}^2$

${\left[(\frac{1}{2}+d)(\frac{1}{2}-d)\right]}^2=\frac{1}{16}$

$[\frac{1}{4}-d^2]=\frac{1}{16}$

$\frac{1}{4}-d^2=\pm \frac{1}{4}$

$d^2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}or{d}^2=0\therefore d=0$ not applicable as$a<b<c$

$d=\pm \frac{1}{\sqrt{2}}$

$\frac{1}{2}-\frac{1}{\sqrt{2}}or\frac{1}{2}+\frac{1}{\sqrt{2}}$