If `|x-1|,3 and |x-3|` are first three terms of an increasing AP, then find the 6th term of on AP .

Explanation

Case I For `xlt1,`
and `|x-3|=-(x-3)`
`therefore 1-x,3" and "3-x` are in AP.
`implies 6=1-x+3-x`
`implies x=-1`
Then, first three terms are `2,3,4,`
which is an increasing AP.
`therefore " 6th "term "is "7." "[because d=1]"`
Case II For `1ltxlt3,`
`|x-1|=x-1`
and `|x-3|=-(x-3)=3-x`
`therefore x-1,3" and "3-x" "are"` in AP.
`implies 6=x-1+3-x`
`implies" " 6=2 " " ["impossible"] `
Case III For `xgt3, |x-1|=x-1` and `|x-3| =x-3`
`therefore x-1,3` and `x-3` are in AP.
`implies 6=x-1+x-3 implies x=5`
Then, first three terms are `4,3,2,` which is a decreasing AP.