Q37 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics

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
128
B
316
Correct answer
C
120
D
312

Solution

$\begin{aligned} & 2 b=a+c, b^2=(a+1)(c+3) \\ & \frac{a+b+c}{3}=8 \rightarrow b=8, a+c=16 \\ & 64=(a+1)(19-a)=19+18 a-a^2 \\ & a^2-18 a-45=0 \rightarrow(a-15)(a+3)=0,(a>10) \\ & a=15, c=1, b=8 \\ & \left((a b c)^{1 / 3}\right)^3=a b c=120\end{aligned}$

Hence, the correct answer is option: 'C'.

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