assume that a=k-d, b=k and c=k+d (since sides r in AP)

 

now, -1<cosB<1 and cosB= a^2+c^2-b^2/2ac(cosine rule)

now, substitute 2b=a+c

 

finally, we get  -2ac< a^2+c^2 < 10/3 ac

since we need exterme values,

equate  10/3 ac=a^2+c^2

 

putting a=k-d and c=k+d, we get a relation btw k and d as k=2d

now, we know b=k and so we get the max value as 2d/3d=2/3

so, the min is obviously 1/3

 

so, the answer is (a)

 

Hi magic....

 

i know that i might not be the right one to provide a soln when there are other talents like shakir,kiran,akriti.....but still i wanted to know if my ans was correct.....so, if i m wrong, pls correct me.

                                                                      goutham harsha