If `a-2b+c=1`, then the value of `|{:(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c):}|` is

If a,b,c are in A.P then the value of det[[x+1,x+2,x+ax+2,x+3,x+bx+3,x+4,x+c]] is

Choose the correct answer in questions 17 to 19: If a, b, c are in A.P., then the determinant [{:(x+2,x+3,x+2a),(x+3,x+4,x+3b),(x+4,x+5,x+2c):}] is : (a) 0 (b) 1 ( c) x (d) 2x

If 1/x+2/3= 1/(2x)+3/4 then the value of x is A) 2 B) 8 C) 6 D) 4

If a,b,c are in AP show that |[x+1,x+2,x+a],[x+2,x+3,x+b],[x+3,x+4,x+c]|=0

if 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

If x-(1)/(x)=1, then the value of x^(3)-(1)/(x^(3)) is equal to (a)2(b)3(c)4(d)8

The value of the determinant |(x, x+a, x+2a),(x,x+2a, x+4a),(x, x+3a, x+6a)| is (A) 0 (B) a^3-x^3 (C) x^3-a^3 (D) (x-a)^3

if (x_(1),x_(2))^(2)+(y_(1)-y_(2))^(2)=a^(2), (x_(2)-x_(3))^(2)+(y_(2)-y_(3))^(2)=b^(2) (x_(3)-x_(1))^(2)+(y_(3)-y_(1))^(2)=c^(2). where a,b,c are positive then prove that 4 |{:(x_(1),,y_(1),,1),(x_(2) ,,y_(2),,1),( x_(3),, y_(3),,1):}| = (a+b+c) (b+c-a) (c+a-b)(a+b-c)