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

• A. 120
• B. 312
• C. 316
• D. 128

Answer 1

Answer: (A)

120

Answer 2

Given that $a, b, c$ are in arithmetic progression, we have:

$2b = a + c$

Since $a + 1, b, c + 3$ are in geometric progression, we know:

$b^2 = (a + 1)(c + 3)$

We are also given that the arithmetic mean of $a, b, c$ is $8$:

$\frac{a + b + c}{3} = 8 \implies a + b + c = 24$

Substituting $b = 8$ (from $a + b + c = 24$), we get $a + c = 16$.

Now, substituting in the geometric progression condition:

$64 = (a + 1)(c + 3) = (a + 1)(19 - a)$

Expanding and rearranging:

$a^2 - 18a + 45 = 0$

Solving this quadratic equation for $a > 10$, we find $a = 15$ and $c = 1$.

Thus, the geometric mean of $a, b, c$ is:

$\sqrt[3]{abc} = \sqrt[3]{15 \times 8 \times 1} = 120$