If a, b and c are three distinct numbers in G | vedclass

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(D) Let the three numbers in $G.P.$ be $a, ar, ar^2$ where $r 
eq 1$ (since the numbers are distinct).Given $a + ar + ar^2 = x(ar)$.Since $a 
eq 0$,

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Let the three numbers in $G.P.$ be $a, ar, ar^2$ where $r 
eq 1$ (since the numbers are distinct).Given $a + ar + ar^2 = x(ar)$.Since $a 
eq 0$,we can divide by $a$:$1 + r + r^2 = xr$$x = \frac{1 + r + r^2}{r} = r + 1 + \frac{1}{r} = (r + \frac{1}{r}) + 1$.We know that for $r > 0$,$r + \frac{1}{r} \geq 2$,so $x \geq 2 + 1 = 3$.For $r  0$. Then $r + \frac{1}{r} = -(k + \frac{1}{k}) \leq -2$.So $x \leq -2 + 1 = -1$.Thus,$x \in (-\infty, -1] \cup [3, \infty)$.Since the numbers are distinct,$r 
eq 1$,so $x 
eq 1 + 1 + 1 = 3$.Therefore,$x$ cannot be any value in the interval $(-1, 3)$.Among the given options,$-2$ is in the range $(-\infty, -1]$,$-3$ is in $(-\infty, -1]$,and $4$ is in $[3, \infty)$.However,$2$ lies in the interval $(-1, 3)$,so $x$ cannot be $2$.

If $a, b$ and $c$ are three distinct numbers in $G.P.$ and $a + b + c = xb$,then $x$ cannot be:

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