Let three real numbers | Sequences and Series| JEE Main 2024

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JEE Main 2024

04-04-2024 S2

Question

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

Select the correct option:

A

120

B

312

C

316

D

128

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

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

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