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

Let d be the common difference of the A.P., then (m−n)d=am−an=1n−1m⇒ d=1mnNow ar−am=(r−m)d⇒ 1=ar=1n+(r−m)1mn=rmn⇒ r=mn.