Given that, mth term=1/n and nth term=1/m.
then ,let a and d be the first term and the common difference of the A.P.

so a+(m-1)d=1/n...........(1)
and a+(n-1)d=1/m...........(2).

subtracting equation (1) by (2) we get,
md-d-nd+d=1/n-1/m

=>d(m-n)=m-n/mn

=>d=1/mn.

again if we put this value in equation (1) or (2) we get, a=1/mn.

then, let A be the mnth term of the AP
a+(mn-1)d=1/mn+1+(-1/mn)=1

proved.
(Q. E. D)

That's all 🙂

Proof:

Let the first term of the AP be 'a' and the common difference be 'd'.

We know that the mth term of the AP is given by:

a + (m-1)d = 1/n ......(1)

And, the nth term of the AP is given by:

a + (n-1)d = 1/M ......(2)

Multiplying equation (1) by 'n' and equation (2) by 'm', we get:

na + (mn - n)d = m/n ......(3)

ma + (mn - m)d = n/M ......(4)

Now, subtracting equation (4) from equation (3), we get:

n(m-1)d - m(n-1)d = m/n - n/M

Simplifying this expression, we get:

d(mn - m - n + 1) = (M - N)/MN

Dividing both sides by (mn - m - n + 1), we get:

d = (M - N)/(MN(mn - m - n + 1)) ......(5)

Substituting this value of 'd' in equation (1), we get:

a = (n - mn)/(mn - m - n + 1) ......(6)

Therefore, the (mn)th term of the AP is given by:

a + (mn - 1)d = [(n - mn)/(mn - m - n + 1)] + [(mn - 1)(M - N)/(MN(mn - m - n + 1))]

Simplifying this expression, we get:

[(mn - 1)MN + n(M - N) - mn(M - N)]/[(mn - m - n + 1)MN]

= (MN - M - N + 1)/[(mn - m - n + 1)MN]

= 1/[(mn - m - n + 1)/MN]

= 1

Hence, the (mn)th term of the AP is 1.