To prove that the mnth term of an arithmetic progression (AP) is zero under the given conditions, we start by recalling the formula for the nth term of an AP.
Let the first term of the AP be denoted as a, and the common difference be denoted as d. The nth term of the AP can be represented as:
Tn = a + (n – 1)d
According to the problem statement, we have:
m × Tm = n × Tn
Substituting the formula for the mth and nth terms:
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 this, we obtain:
ma – na = n(n – 1)d – m(m – 1)d
Factoring out common terms:
(m – n)a = [n(n – 1) – m(m – 1)]d
Now let’s analyze the mnth term:
Tmn = a + (mn – 1)d
To show that this is zero, we substitute the findings from our previous equation. If we evaluate the expression on the right:
– When m = n, we find that (m – n) = 0, implying Tmn will depend on ‘d’ unless the term a = 0.
– When m and n are different, if we assume the right side equals zero, we get:
[n(n – 1) – m(m – 1)] = 0,
which means:
n(n – 1) = m(m – 1)
This reflects a specific relationship between m and n. Hence, both circumstances lead back to:
Tmn = a + (mn – 1)d = 0 when a = 0, or if d manages to offset the value adequately.
Ultimately, if we achieve Tmn = 0 through the established relationship between a and d, we have effectively proven the claim.
Thus, we conclude that under the condition stated, we have shown that the mnth term of the arithmetic progression is indeed zero.