Suppose a , b , c are in A.P. and a ², 2 ~b ², c ² are in G.P. If a < b < c and a + b + c =1 , then 9 ( a ²+ b ²+ c ² ) is equal to _ _ _ _ .

Step-by-step solution

Since a, b, c are in A.P., we have 2b = a + c . With a + b + c = 1 , we get 3b = 1 , so b = 1 3 and a + c = 2 3 . Since a^2, 2b^2, c^2 are in G.P.: (2b^2)^2 = a^2 c^2 , giving 4b^4 = a^2c^2 . With b = 1 3 : a^2c^2 = 4 81 . For real values with a Thus a, c satisfy t^2 - 2 3 t - 2 9 = 0 , giving t = 1 3 3 . So a = 1- 3 3 , c = 1+ 3 3 . Calculating: a^2 + c^2 = (1- 3 )^2 + (1+ 3 )^2 9 = 4-2 3 +4+2 3 9 = 8 9 . Therefore a^2 + b^2 + c^2 = 8 9 + 1 9 = 1 , and 9(a^2 + b^2 + c^2) = 9 .