Let three real numbers a, b, c be in arithmetic progression and a+1, b, c+3 be in geometric progression. If a10 and the arithmetic mean of a, b and c is 8, then the cube of the geometric mean of a, b and c is

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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

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Answer: A. 120 \begin{aligned} & 2 b=a+c \quad 	ext{.... (1)}\ & b^2=(a+1)(c+3) \quad 	ext{.... (2)}\ & \frac{a+b+c}{3}=8 \quad 	ext{.... (3)} \end{aligned} \begin{aligned} \Rightarrow & \frac{3 b}{3}=8 \ & b=8 \ \Rightarrow \quad & a c+3 a+c+3=64 \end{aligned} \begin{aligned} & 3 a+c+a c=61 \quad 	ext{... (4)}\ & a+c=16 \ & c=16-a \end{aligned} from equation (4) \begin{aligned} & 3 a+16-a+a(16-a)=61 \ & \Rightarrow \quad(a-15)(a-3)=0 \ & \quad a=15(a>10) \ & \Rightarrow \quad a=15, b=8, c=1 \ & \left((a \cdot b \cdot c)^{\frac{1}{3}}ight)^3=15 	imes 8 	imes 1=120 \end{aligned}

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

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