If mth term of an AP is 1/n and its nth term is 1/m , then show that its (mn)th term is 1

Explanation

Let the first term and the common difference of the A.P. be 'a' and 'd' respectively.
`:. a_(m)=(1)/(n) rArr a+(m-1)d=(1)/(n) " " ...(1)`
and `a_(n)=(1)/(m) rArr a+(n-1)d=(1)/(m) " " ...(2)`
Subtract equation (2) from equation (1), we get
`{:(" "a+(m-1)d=""(1)/(n)),(underset(-)" "aunderset(-)+(n-1)d=underset(-)""(1)/(m)),( bar(" "(m-n)d=(1)/(n)-(1)/(m))),():}`
`rArr (m-n)d=(m-n)/(nm)`
`rArr d=(1)/(mn)`
put `d=(1)/(mn)` in equations (1), we get
`a+(m-1).(1)/(mn)=(1)/(n)`
`rArr a+(1)/(n)-(1)/(mn)=(1)/(n)`
`rArr a=(1)/(mn)`
Now, `a_(mn)=a+(mn-1)d`
`=(1)/(mn)+(mn-1).(1)/(mn)=(1)/(mn)+1-(1)/(mn)=1`
`:. (mn)`th term of the A.P.=1 `" "`Hence Proved.