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
- 120
- 312
- 316
- 128
The Correct Option is A
Solution and Explanation
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:
\[ \sqrt[3]{abc} = \sqrt[3]{15 \times 8 \times 1} = 120 \]
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