Suppose m th term of an A.P. is 1/n and nth term of the A.P. is 1/m. If rth term of the A.P. is 1, then r is equal to
mn
m+n
m–n
m+n –1
Let d be the common difference of the A.P., then
(m−n)d=am−an=1n−1m
⇒ d=1mn
Now ar−am=(r−m)d
⇒ 1=ar=1n+(r−m)1mn=rmn⇒ r=mn.