Sequences and Series 3 Question 9
9. Let $a, b, c$ be in an AP and $a^{2}, b^{2}, c^{2}$ be in GP. If $a<b<c$ and $a+b+c=\frac{3}{2}$, then the value of $a$ is
(2002, 1M)
(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: 9. (a)
Solution:
- Since, $a, b$ and $c$ are in an AP.
Let $\quad a=A-D, b=A, c=A+D$
Given, $\quad a+b+c=\frac{3}{2}$
$\Rightarrow \quad(A-D)+A+(A+D)=\frac{3}{2}$
$\Rightarrow \quad 3 A=\frac{3}{2} \Rightarrow A=\frac{1}{2}$
$\therefore$ The number are $\frac{1}{2}-D, \frac{1}{2}, \frac{1}{2}+D$.
Also, $\frac{1}{2}-D^{2}, \frac{1}{4}, \frac{1}{2}+D^{2}$ are in GP.
$\therefore \quad \frac{1}{4}{ }^{2}=\frac{1}{2}-D^{2} \frac{1}{2}+D^{2} \Rightarrow \frac{1}{16}=\frac{1}{4}-D^{2}$
$\Rightarrow \quad \frac{1}{4}-D^{2}= \pm \frac{1}{4} \Rightarrow D^{2}=\frac{1}{2} \Rightarrow D= \pm \frac{1}{\sqrt{2}}$
$\therefore \quad a=\frac{1}{2} \pm \frac{1}{\sqrt{2}}$
So, out of the given values, $a=\frac{1}{2}-\frac{1}{\sqrt{2}}$ is the right choice.
