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`

Answer

N/a

Explanation

True
Since `a,b` and `c` are in AP then `2b=a+c`
`:. |(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)|=0`
`implies |(2x+4,2x+6,2x+a+c),(x+2,x+3,x+b),(x+3,x+4,x+c)|=0 [ :' R_(1)toR_(1)+R_(3)]`
`implies |(2(x+2),2(x+3),2(x+b)),(x+2,x+3,x+b),(x+3,x+4,x+c)|=0 [ :' 2b=a+c]`
`implies 0=0` [since `R_(1)` and `R_(2)` are in proportional to each other]
Hence statement is true.