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(D) Let the three distinct numbers in $G.P.$ be $a, ar, ar^2$ where $r 
eq 1$ and $r 
eq 0$.Given $a + b + c = xb$,substituting the terms: $a + ar +

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$2$

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(D) Let the three distinct numbers in $G.P.$ be $a, ar, ar^2$ where $r 
eq 1$ and $r 
eq 0$.Given $a + b + c = xb$,substituting the terms: $a + ar + ar^2 = x(ar)$.Since $a 
eq 0$,we can divide by $a$: $1 + r + r^2 = xr$.Dividing by $r$: $r + 1 + \frac{1}{r} = x$,which simplifies to $x = r + \frac{1}{r} + 1$.We know that for any real $r > 0$,$r + \frac{1}{r} \geq 2$,and for $r Since the numbers are distinct,$r 
eq 1$ and $r 
eq -1$.If $r > 0$ and $r 
eq 1$,then $r + \frac{1}{r} > 2$,so $x > 3$.If $r Thus,$x$ cannot lie in the interval $(-1, 3)$.Among the given options,$2$ 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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