Question

If a, b and c are three distinct real numbers in G.P and $a+b+c=xb$, then the value of x cannot be
(a) 4
(b) –3
(c) –2
(d) 2

Answer 1

We start solving the problem by recalling the general form of terms in a G.P (Geometric Progression). We assume a common ratio as ‘r’ and write all the given numbers in terms of ‘a’ and ‘r’. After writing we get a quadratic equation in r after making subsequent arrangements. We take discriminant greater than zero to get the solution set for x.

Complete step-by-step answer:
According to the problem, we have three distinct numbers a, b and c which are in G.P (Geometric Progression) and these numbers satisfy $a+b+c=xb$. We need to find the range of x so that the values that are not valid for x can be chosen.
We know that the general terms of the G.P (Geometric Progression) is defined as $p$, $pr$, $p{{r}^{2}}$,…….,$p{{r}^{n-1}}$. Where r is a common ratio. We apply this definition to the numbers a, b and c.
We get $b=ar$ and $c=a{{r}^{2}}$. We substitute these values in $a+b+c=xb$.
$\Rightarrow a+ar+a{{r}^{2}}=x\left( ar \right)$.
$\Rightarrow a+ar-xar+a{{r}^{2}}=0$.
$\Rightarrow a+ar\left( 1-x \right)+a{{r}^{2}}=0$.
$\Rightarrow 1+r\left( 1-x \right)+{{r}^{2}}=0$.
$\Rightarrow {{r}^{2}}+r\left( 1-x \right)+1=0$ -(1).
We have given that a, b and c as three distinct numbers. So, we have three distinct real values of a, b and c. If we cannot have a solution for equation (1), then a, b and c cannot be distinct values.
So, equation (1) must have real roots for ‘r’ in order to get the distinct values for a, b and c.
We know that to get real for the quadratic equation $a{{x}^{2}}+bx+c=0$, we should have discriminant (D) greater than 0 i.e., $D=\sqrt{{{b}^{2}}-4ac}>0$.
So, take determinant for equation (1).
$\Rightarrow {{\left( 1-x \right)}^{2}}-4\left( 1 \right)\left( 1 \right)>0$.
$\Rightarrow {{x}^{2}}-2x+1-4>0$.
$\Rightarrow {{x}^{2}}-2x-3>0$.
$\Rightarrow {{x}^{2}}-3x+x-3>0$.
$\Rightarrow x.\left( x-3 \right)+1.\left( x-3 \right)>0$.
$\Rightarrow \left( x+1 \right)\left( x-3 \right)>0$ -(2).
We know that the solution set for x in the inequality $\left( x-a \right).\left( x-b \right)\ge 0$ and $a < b$ is $x < a \text{ or }x > b$.
We get the solution set for x of the inequality in equation (2) is $x < -1\text{ or }x > 3$.
From options we can see that 2 is not in the solution set $x < -1\text{ or } x > 3$.
x cannot be equal to 2.

So, the correct answer is “Option d”.

Note: If we take real and equal root conditions for equation (1), we get as follows.
$\Rightarrow D=\sqrt{{{b}^{2}}-4ac}=0$.
$\Rightarrow {{\left( 1-x \right)}^{2}}-4\left( 1 \right)\left( 1 \right)=0$.
$\Rightarrow {{x}^{2}}-2x+1-4=0$.
$\Rightarrow {{x}^{2}}-2x-3=0$.
$\Rightarrow {{x}^{2}}-3x+x-3=0$.
$\Rightarrow x.\left( x-3 \right)+1.\left( x-3 \right)=0$.
$\Rightarrow \left( x+1 \right)\left( x-3 \right)=0$.
$\Rightarrow \left( x+1 \right)=0\text{ or }\left( x-3 \right)=0$.
$\Rightarrow x=-1\text{ or }x=3$.
Let us substitute $x=-1$ in equation (1).
$\Rightarrow {{r}^{2}}+r\left( 1-\left( -1 \right) \right)+1=0$.
$\Rightarrow {{r}^{2}}+r\left( 1+1 \right)+1=0$.
$\Rightarrow {{r}^{2}}+2r+1=0$.
$\Rightarrow {{\left( r+1 \right)}^{2}}=0$.
$\Rightarrow r+1=0$.
$\Rightarrow r=-1$.
We can see that $c=a{{\left( -1 \right)}^{2}}$.
$\Rightarrow c=a\left( 1 \right)$.
$\Rightarrow c=a$, but we need all the numbers a, b and c distinct. So, x cannot be equal to -1.
Let us substitute $x=3$ in equation (1).
$\Rightarrow {{r}^{2}}+r\left( 1-3 \right)+1=0$.
$\Rightarrow {{r}^{2}}+r\left( -2 \right)+1=0$.
$\Rightarrow {{r}^{2}}-2r+1=0$.
$\Rightarrow {{\left( r-1 \right)}^{2}}=0$.
$\Rightarrow r-1=0$.
$\Rightarrow r=1$.
We can see that $c=a{{\left( 1 \right)}^{2}}$.
$\Rightarrow c=a\left( 1 \right)$.
$\Rightarrow c=a$, but we need all the numbers a, b and c distinct. So, x cannot be equal to 3.