Question 63

Let three real numbers $$a, b, c$$ be in arithmetic progression and $$a + 1, b, c + 3$$ be in geometric progression. If $$a > 10$$ and the arithmetic mean of $$a, b$$ and $$c$$ is $$8$$, then the cube of the geometric mean of $$a, b$$ and $$c$$ is

a,b,c in AP: b=8(AM). So a+b+c=24, b=8. a+c=16, c=16-a.

a+1,b,c+3 in GP: b²=(a+1)(c+3). 64=(a+1)(19-a).

64=19a-a²+19-a=18a-a²+19. a²-18a+45=0. a=(18±√(324-180))/2=(18±12)/2.

a=15 or a=3. Since a>10: a=15,b=8,c=1.

GM=(abc)^{1/3}=(15·8·1)^{1/3}=(120)^{1/3}. GM³=120.

The answer is Option (3): 120.

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