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 [2024]

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 [2024]

Smart Achievers · Accepted Answer

(3)

$2 b = a + c$ $[∵ a, b, c$ are in A.P. $]$ ...(i)

$b^{2} = (a + 1) (c + 3)$ $[∵ a + 1, b, c + 3$ are in G.P. $]$ ...(ii)

Also, $\frac{a + b + c}{3} = 8$ $[∵$ A.M. of $a, b, c]$ ...(iii)

$\Rightarrow \frac{3 b}{3} = 8 \Rightarrow b = 8$ [Using (i)]

By (ii), $64 = (a + 1) (c + 3)$ and by (iii), $c = 16 - a$

$\Rightarrow 64 = (a + 1) (16 - a + 3)$

$\Rightarrow a^{2} - 18 a + 45 = 0 \Rightarrow a = 15, 3$

$∴$ $a = 15$ $(a \neq 3, as a > 10)$

Also, $c = 1$

$∴$ ${[{(a \cdot b \cdot c)}^{1 / 3}]}^{3} = 15 \times 8 \times 1 = 120$

Solution

Q.
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 [2024]

1 316

2 128

3 120

4 312

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