Sequences and Series 3 Question 6
6. If $a, b$ and $c$ be three distinct real numbers in GP and $a+b+c=x b$, then $x$ cannot be
(2019 Main, 9 Jan I)
(a) 4
(b) 2
(c) -2
(d) -3
Show Answer
Answer:
Correct Answer: 6. (b)
Solution:
- Let $b=a r$ and $c=a r^{2}$, where $r$ is the common ratio.
We know that, $r+\frac{1}{r} \geq 2($ for $r>0)$
and $\quad r+\frac{1}{r} \leq-2($ for $r<0)[$ using $AM \geq GM]$
$$ \begin{array}{lc} \therefore & 1+r+\frac{1}{r} \geq 3 \\ \text { or } & 1+r+\frac{1}{r} \leq-1 \\ \Rightarrow & x \geq 3 \text { or } x \leq-1 \\ \Rightarrow & x \in(-\infty,-1] \cup[3, \infty) \end{array} $$
Hence, $x$ cannot be 2 .
Alternate Method
From Eq. (i), we have
$$ \Rightarrow \quad \begin{array}{r} 1+r+r^{2}=x r \\ \quad r^{2}+(1-x) r+1=0 \end{array} $$
For real solution of $r, D \geq 0$.
$$ \begin{array}{ll} \Rightarrow & (1-x)^{2}-4 \geq 0 \\ \Rightarrow & x^{2}-2 x-3 \geq 0 \\ \Rightarrow & (x-3)(x+1) \geq 0 \\ & \quad+\quad-\quad+ \\ \hline-1 & 3 \end{array} $$
$\Rightarrow x \in(-\infty,-1] \cup[3, \infty)$
$\therefore x$ cannot be 2 .
