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If a,b,c, are in A.P. then |(x+1,x+2,x+a...

If a,b,c, are in A.P. then `|(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)|=`

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If a,b,c are A.P then |{:(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

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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

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Given a,b,c are in A.P. Then determinant : |(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)| in its simplified form is :

If a,b,c and in A.P, then x^a,x^b,x^c are in:

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