If the m th term of an Arithmetic Progression (AP) is 1/n and n th term is 1/m, then find the sum of mn terms.

let us assume a be the first term and d the common difference.
According to Question,
m^th term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for m^th term,
⇒ a + ( m - 1 ) x d = t_m
⇒ a + ( m - 1 ) x d = 1/n
⇒ a + md - d = 1/n .......................(1)
Use the Arithmetic Progression formula for n^th term,
⇒ a + ( n - 1 ) x d = t_n
⇒ a + ( n - 1 ) x d = 1/m
⇒ a + nd - d = 1/m.........................(2)
Subtracts Equation (1) from Equation (2) , we will get
⇒ a + nd - d - (a + md - d) = 1/m - 1/n
⇒ a + nd - d - (a + md - d) = 1/m - 1/n
⇒ a + nd - d - a - md + d = 1/m - 1/n
⇒ nd - md = 1/m - 1/n
⇒ d(n - m) = 1/m - 1/n
⇒ d(n - m) = (n - m)/mn
⇒ d = 1/mn.......................................(3)
Put the value of d in Equation (1), we will get,
⇒ a + md - d = 1/n
⇒ a + ( m x 1/mn ) - 1/mn = 1/n
⇒ a + 1/n - 1/mn = 1/n
⇒ a = 1/n - 1/n + 1/mn
⇒ a = 1/mn.....................................(4)
Now according to question,
Sum of mn terms = mn/2(2a + ( mn - 1) x d)
S_mn = mn/2(2a + ( mn - 1) x d)
Put the value of a and d in above equation, we will get,
S_mn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ]
S_mn = mn/2 [2/mn + 1 - 1/mn ]
S_mn = mn/2 [ 1 + 1/mn ]
S_mn = mn/2 [ ( mn + 1 )/mn ]
S_mn = mn/2 (1 + mn)/mn
S_mn = 1/2 (1 + mn)
S_mn = (1 + mn)/2 = (mn + 1)/2