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Let a, b and c be distinct real numbers. If a, b, c are in geometric progression and a + b + c = xb, then x lies in the set

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a
(1, 3)
b
(–1, 0) ∪ (1, 2)
c
(−∞,−1)∪(3,∞)
d
(0,1)

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detailed solution

Correct option is C

a+b+c=xb⇒br+b+br=xb, where r is commonratio of the G.P and r ≠ 1⇒r+1r=(x−1)But r+1r<−2 or r+1r>2∴x−1<−2 or x−1>2⇒x<−1 or x>3⇒x∈(−∞,−1)∪(3,∞)


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