Q91 / 200 · Manipal (MET) 2012 · Mathematics
Suppose $a, b, c$ are in AP and $a^2, b^2, c^2$ are in GP. If $a \lt b \lt c$ and $a+b+c=\frac{3}{2}$, then the value of $a$ is
- A$\frac{1}{2 \sqrt{2}}$
- B$\frac{1}{2 \sqrt{3}}$
- C$\frac{1}{2}-\frac{1}{\sqrt{3}}$
- Correct answerD$\frac{1}{2}-\frac{1}{\sqrt{2}}$
Solution
Since, $a, b, c$ are in AP.
$\therefore \quad a=A-D, b=A, c=A+D$
Where, $A$ is the first term and $D$ is the common difference of an AP.
Given,
$a+b+c=\frac{3}{2}$
$\begin{array}{rrr}
\Rightarrow & (A-D)+A+(A+D) & =\frac{3}{2} \\
\Rightarrow & 3 A & =\frac{3}{2} \\
\Rightarrow & A & =\frac{1}{2}
\end{array}$
$\therefore$ 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.
$\Rightarrow \quad\left(\frac{1}{4}\right)^2=\left(\frac{1}{2}-D\right)^2\left(\frac{1}{2}+D\right)^2$
$\begin{array}{ll}
\Rightarrow & \frac{1}{16}=\left(\frac{1}{4}-D^2\right)^2 \\
\Rightarrow & \frac{1}{4}-D^2= \pm \frac{1}{4} \\
\Rightarrow & \quad D^2=\left(\frac{1}{2} D=0 \text { is not possible }\right) \\
\Rightarrow & D= \pm \frac{1}{\sqrt{2}} \\
\therefore & a=\frac{1}{2} \pm \frac{1}{\sqrt{2}}
\end{array}$
So, out of the given value $a=\frac{1}{2}-\frac{1}{\sqrt{2}}$ is the right choice.
Hence, the correct answer is option: 'D'.