1. If mth term is 1/n and nth term is 1/m in an A.P, prove that the sum of first (mn) terms of the series is [mn+1]/2?

Suppose a_k = a + ( k - 1 )d , then
a_m = a + ( m - 1 )d = 1/n ..... (1)
a_n = a + ( n - 1 )d = 1/m ..... (2)

(2) - (1)
( n - m )d = 1/m - 1/n = ( n - m )/( mn )
d = 1/( mn ) , substitute into (1)
a = 1/n - ( m - 1 )/( mn ) = 1/n - 1/n + 1/( mn ) = 1/( mn )

Hence,
a_k = 1/( mn ) + ( k - 1 )/( mn ) = k / ( mn )
a_mn = mn / ( mn ) = 1

S
= mn * ( a_1 + a_mn ) / 2
= mn * [ 1/( mn ) + 1 ] / 2
= ( 1 + mn ) / 2

Q.E.D.