Suppose a, b, c are in AP and a^2, b^2, c^2 are in GP. If a b c and a+b+c= 3 2 , then the value of a is

Step-by-step solution

Since, a, b, c are in AP. a=A-D, b=A, c=A+D Where, A is the first term and D is the common difference of an AP. Given, a+b+c= 3 2 array rrr & (A-D)+A+(A+D) & = 3 2 & 3 A & = 3 2 & A & = 1 2 array The numbers are 1 2 -D, 1 2 , 1 2 +D Also, ( 1 2 -D )^2, 1 4 , ( 1 2 +D )^2 are in GP. ( 1 4 )^2= ( 1 2 -D )^2 ( 1 2 +D )^2 array ll & 1 16 = ( 1 4 -D^2 )^2 & 1 4 -D^2= 1 4 & D^2= ( 1 2 D=0 is not possible ) & D= 1 2 & a= 1 2 1 2 array So, out of the given value a= 1 2 - 1 2 is the right choice.