Question

If $a,b,c$ are three distinct real numbers in G.P. and $a+b+c=xb$, then prove that either $x < -1$ or $x > 3$.

Answer 1

Start solving this question by assuming that $a$ is the first term of G.P., $b$ is the second term and $c$ is the third term and $r$ be the common ratio of G.P.. Now, calculate the values of terms and substitute in the given expression $a+b+c=xb$ to obtain a desired answer.

Complete step by step answer:
We have been given that $a,b,c$ are three distinct real numbers in G.P. and $a+b+c=xb$.
We have to prove that either $x < -1$ or $x > 3$.
Now, let us assume that $a$ is the first term of G.P., $b$ is the second term of G.P. and $c$ is the third term of G.P.
Also let us assume that $r$ be the common ratio of G.P.
Now, we know that if $a$ is the first term of G.P. and $r$ is the common ratio of G.P. then second and third term will be
$\begin{aligned} & b=ar \\\ & c=a{{r}^{2}} \\\ \end{aligned}$
Now, we have given $a+b+c=xb$.
Substituting the values of $a,b,c$, we get
$\begin{aligned} & a+ar+a{{r}^{2}}=xar \\\ & \Rightarrow a{{r}^{2}}-xar+ar+a=0 \\\ \end{aligned}$
Let us divide the whole equation by $a$, we get
$\begin{aligned} & \Rightarrow \dfrac{a{{r}^{2}}}{a}-\dfrac{xar}{a}+\dfrac{ar}{a}+\dfrac{a}{a}=0 \\\ & \Rightarrow {{r}^{2}}-xr+r+1=0 \\\ \end{aligned}$
Now, factorizing the above equation, we get
$\Rightarrow {{r}^{2}}+r\left( 1-x \right)+1=0$
The above equation is a quadratic equation and as given in the question $a,b,c$ are real numbers, so the value of discriminant of equation is always greater than or equal to zero.
$D\ge 0$, Where $D={{b}^{2}}-4ac$
Now, substituting the values, we get

& D={{\left( 1-x \right)}^{2}}-4.1.1 \\\ & {{\left( 1-x \right)}^{2}}-4.1.1\ge 0 \\\ & {{\left( 1-x \right)}^{2}}-4\ge 0 \\\ & {{x}^{2}}+1-2x-4\ge 0 \\\ & {{x}^{2}}-2x-3\ge 0 \\\ & {{x}^{2}}-3x+x-3\ge 0 \\\ & x\left( x-3 \right)+1\left( x-3 \right)\ge 0 \\\ & \left( x-3 \right)\left( x+1 \right)\ge 0 \\\ \end{aligned}$$ So, $x < -1$ or $x > 3$. Hence proved **Note:** In geometric progression each term after first term is found by multiplying the previous term by a number called common ratio. Discriminant is an expression which is used to determine how many real solutions the quadratic equation has. The possibility of mistake is that if students take $$D=0$$ it means the quadratic equation has one solution.