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
- 316
- 312
- 128
Answer
(1) 120
Solution
A.P.: 2b = a + c; A.M. = 8 a + b + c = 24
3b = 24 b = 8, a + c = 16
G.P.: b2 = (a + 1)(c + 3) 64 = (a + 1)(19 - a)
a2 - 18a + 45 = 0 (a - 15)(a - 3) = 0
a > 10 a = 15, c = 1
(G.M.)3 = abc = 15 8 1 = 120
Answer: (1)