Sequences And Series Ques 74
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: 74.(b)
Solution:
Formula:
- Let $b=a r$ and $c=a r^{2}$, where $r$ is the common ratio.
Then , $a + b + c = xb$
$\Rightarrow a+ ar + ar^2 =xar$
$\Rightarrow 1+ r + r^2 = xr $ $\quad$ …….(i)
$\Rightarrow x = \frac{1+ r + r^2}{r}= 1+r+ \frac{1}{r}$
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 $r, D \geq 0$.
$\Rightarrow (1-x)^{2}-4 \geq 0 $
$\Rightarrow x^{2}-2 x-3 \geq 0 $
$\Rightarrow (x-3)(x+1) \geq 0 $
$\Rightarrow x \in(-\infty,-1] \cup[3, \infty)$
$\therefore x$ cannot be $ 2 $.
BETA
