Question

Q.8

Answer 1

Let 'a' be the first term and 'd' be the common difference of given A.P 

`m^(th) "term" =1/n`

`⇒ a+(m-1)d=1/n .................(1)`

`n^(th) "term" =1/m` 

`⇒a(n-1)d=1/m...............(2)`

Subtracting (2) from(1), we get 

`(m-1)d-(n-1)d=1/n-1/m`

`⇒md-d-nd+d=(m-n)/(mn)`

`⇒(m-n)d=(m-n)/(mn)`

`⇒ d=1/(mn)`

Substituting value of d in (1),we get 

`a+(m-1)xx1/(mn)=1/n`

`⇒ a=1/n-(m-1)/(mn)=(m-m+1)/(mn)=1/(mn)` 

Now, 

`(mn)^(th) "trem"=a+(mn-1)d`

                          `=1/(mn)+(mn-1)xx1/(mn)`

                          ` =(1+mn-1)/(mn)`

                         `=(mn)/(mn)`

                        ` =1`