If the `m^(t h)` term of an A.P. is `1/n a n d\ t h e\ n^(t h)t e r m\ i s1/m ,\ ` show that the sumof `m n` terms is `1/2(m n+1)` where `m!=ndot`

Explanation

Let sum be a

SO,


`a_{m}=a+(m-1) d=1 / n ldots ldots . .(1)`


nth term `=1 / {m}`


`a_{n}=a+(n-1) d=1 / m ldots ldots . .(2)`
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