IIT JEE , AIEEE Forum - Mathematics , Trignometry

kghedriu

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)

hey, rate me if my ans is correct