Answer :

Let a be the first term of GP with r being the common ratio.


b = ar …(1)


c = ar2 …(2)


Given,


(a + b + c) = xb


(a + ar + ar2) = x(ar)


a(1 + r + r2) = ar


(1 + r + r2) = xr


r2 + (1 – x)r + 1 = 0


As r is a real number Both solutions are real.


So discriminant of the given quadratic equation D ≥ 0


As, D ≥ 0


(1 – x)2 – 4(1)(1) ≥ 0


x2 – 2x – 3 ≥ 0


(x – 1)(x – 3) ≥ 0


x < – 1 or x > 3 …proved


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