Show that `|x+1x+2x+a x+2x+3x+b x+3x+4x+c|=0` where `a ,\ b ,\ c` are in A.P.

Show that |[x+1,x+2,x+a], [x+2,x+3,x+b],[ x+3,x+4,x+c]|=0 where a ,\ b ,\ c are in A.P.

|{:(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c):}|=0 where a,b,c are in A.P

Without expanding,show that abs((x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)) =0,Where a ,b,c are in A.P

If ||(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)|=0 then a,b,c are

If the lines 2x - ay + 1 = 0, 3x – by + 1 = 0, 4x-cy+1 =0 concurrent, then a, b, c are in (A) A.P. (B) G.P. (C) H.P. (D) A.G.P

Let |[x^2+3x,x-1,x+3],[x+1,-2x,x-4],[x-3,x+4, 3x]|=a x^4+b x^3+c x^2+e be an identity in x , where a , b , c , d , e are independent of xdot Then the value of e is (a) 4 (b) 0 (c) 1 (d) none of these