If a, b, c are in A.P. and a^2, b^2, c^2 are in G.P. such that a < b < c and a+b+c= 3 4 , then the value of a is

Step-by-step solution

a, b, c are in A.P. then a+c=2 b also it is given that, aligned &a+b+c= 3 4 & 2 b+b= 3 4 b= 1 4 aligned Again it is given that, a^2, b^2, c^2 are in G.P. then (b^2 )^2=a^2 c^2 a c= 1 16 From (1), (2) and (3), we get; a 1 16 a = 1 2 16 a^2-8 a 1=0 Case I: 16 a^2-8 a+1=0 a= 1 4 (not possible as .a < b ) Case II: 16 a^2-8 a-1=0 a= 8 128 32 aligned & a= 1 4 1 2 2 & a= 1 4 - 1 2 2 ( a < b) aligned