If `1/7=0. 142857 ` , write the decimal expression of `2/7,3/7,4/7,5/7a n d6/7` without actually doing the long division.

Explanation

i) `2/7`
`2/7=2xx(1/7)`
`=2xx0.bar142857`
`=0.bar285714`
Also, we observe that we get `2` as a remainder after the second step in the above division
Hence, we start writing the quotient after the second decimal place and we get `2/7=0.bar285714`
Hence, `2/7=0.285714`
ii) `3/7`
`3/7=3xx(1/7)`
`=3xx0.bar142857`
`=0.bar428571`
Also, we observe that we get `3` as a remainder after the first step in the above division
Hence, we start writing the quotient after the first decimal place and we get `3/7=0.bar428571`
Hence, `3/7=0.bar428571`
iii) `4/7`
`4/7= 4xx(1/7)`
`=4xx0.bar142857`
`=0.bar571428`
Also, we observe that we get `4` as a remainder after the fourth step in the above division
Hence, we start writing the quotient after the fourth decimal place and we get `4/7=0.571428`
Hence, `4/7=0.bar571428`
iv) `5/7`
`5/7=5xx(1/7)`
`=5xx0.142857`
`=0.bar714285`
Also, we observe that we get `5` as a remainder after the fifth step in the above division
Hence, we start writing the quotient after the fifth decimal place and we get `5/7=0.bar714285`
Hence, `5/7=0.bar714285`
v) `6/7`
`6/7=6xx(1/7)`
`=6xx0.bar142857`
`=0.bar857142`
Also, we observe that we get `6` as a remainder after the third step in the above division
Hence, we start writing the quotient after the third decimal place and we get `6/7=0.bar857142`
Hence, `6/7=0.bar857142`