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Q.
Suppose $a,\, b,\, c$ are in $AP$ and $a^{2},\, b^{2},\, c^{2}$ are in GP. If $a < b < c$ and $a+b+c=\frac{3}{2}$, then the value of $a$ is
ManipalManipal 2012
Solution:
Since, $a, b, c$ are in $AP$.
$\therefore a=A-D,\, b=A,\, c=A+D$
Where, $A$ is the first term and $D$ is the common difference of an AP.
Given, $a + b +c=\frac{3}{2}$
$\Rightarrow (A-D)+A+(A+D)=\frac{3}{2}$
$\Rightarrow 3 A=\frac{3}{2}$
$\Rightarrow A=\frac{1}{2}$
$\therefore $ The numbers are $\frac{1}{2}-D,\, \frac{1}{2}, \frac{1}{2}+D$
Also, $\left(\frac{1}{2}-D\right)^{2}, \frac{1}{4},\left(\frac{1}{2}+D\right)^{2}$ are in GP.
$\Rightarrow \left(\frac{1}{4}\right)^{2}=\left(\frac{1}{2}-D\right)^{2}\left(\frac{1}{2}+D\right)^{2}$
$\Rightarrow \frac{1}{16}=\left(\frac{1}{4}-D^{2}\right)^{2}$
$\Rightarrow \frac{1}{4}-D^{2}=\pm \frac{1}{4}$
$\Rightarrow D^{2}=\left(\frac{1}{2} D=0\right.$ is not possible)
$\Rightarrow D=\pm \frac{1}{\sqrt{2}}$
$\therefore a=\frac{1}{2} \pm \frac{1}{\sqrt{2}}$
So, out of the given value $a=\frac{1}{2}-\frac{1}{\sqrt{2}}$ is the right choice.