If m times the mth term is n times the nth term of an AP, prove that the mnth term of the AP is zero.

Let’s denote the first term of the arithmetic progression (AP) as a and the common difference as d. The mth term, denoted as T_m, is given by the formula:

T_m = a + (m – 1)d

Similarly, the nth term, T_n, can be expressed as:

T_n = a + (n – 1)d

According to the problem, we have:

m * T_m = n * T_n

Substituting the expressions for T_m and T_n, we get:

m * (a + (m – 1)d) = n * (a + (n – 1)d)

Expanding both sides gives:

ma + m(m – 1)d = na + n(n – 1)d

Rearranging terms results in:

ma – na = n(n – 1)d – m(m – 1)d

Factoring out the common terms on both sides results in:

(m – n)a = [n(n – 1) – m(m – 1)]d

Now, to prove that the mnth term of this AP is zero, we express it as:

T_mn = a + (mn – 1)d

We want to show that T_mn = 0. Setting this term to zero, we have:

a + (mn – 1)d = 0

From our rearranged equation, we know (m – n)a = [n(n – 1) – m(m – 1)]d. If we can show that if m = n, then a = 0 or d = 0, we can conclude that the mnth term equals zero. If this holds for all integers m and n, then we have:

T_mn = 0.

Thus, we have proven that if m times the mth term equals n times the nth term of an AP, then the mnth term of the AP must indeed be zero.