Question

4k + 8, 2k 2 + 3k + 6, 3k 2 + 4k + 4 are the angles of a triangle. These form an A.P. Find value of k.

Answer 1

4k+8....
2k^2+3k+6....
3k^2+4k+4

(3k^2+4k+4) -(2k^2+3k+6) =(2k^2+3k+6) -(4k+8)
k^2+k-2=2k^2-k-2
k^2-2k=0
k(k-2) =0
thus....
k=0 or k= 2
if k=2....
angles are 16, 20, 24...sum =60...
3 times less... so angles are...
16*3; 20*3; 24*3..
48; 60; 72....
48=4k+8
k=10