If p times the pᵗʰ term of an AP be equal to q times the qᵗʰ term, then (p +…

Concept

In an arithmetic progression (AP) with first term a and common difference d, the nth term is given by the formula tn = a + (n − 1)d. A single linear equation in the two unknowns a and d cannot fix both values, but it can force one particular term to vanish — that is the idea here.

Application

Write the two terms named in the question using the formula:

1. pth term = a + (p − 1)d, and qth term = a + (q − 1)d.

2. Translate the condition “p times the pth term equals q times the qth term”: p[a + (p − 1)d] = q[a + (q − 1)d].

3. Bring everything to one side and group the a-terms and the d-terms: (p − q)a + [p(p − 1) − q(q − 1)]d = 0.

4. Factor the d-coefficient. Since p(p − 1) − q(q − 1) = p2 − p − q2 + q = (p − q)(p + q) − (p − q) = (p − q)(p + q − 1), the whole equation becomes (p − q)[a + (p + q − 1)d] = 0.

5. Because p and q are different, the factor (p − q) is non-zero, so it cancels and we are left with a + (p + q − 1)d = 0.

6. Now read the left side as a term: a + (p + q − 1)d is precisely the (p + q)th term of the same AP. Hence that term equals 0.

Cross-check

Take a concrete AP that satisfies the condition, e.g. p = 2, q = 3, and pick a, d so that 2·t2 = 3·t3. Solving a + (2 + 3 − 1)d = a + 4d = 0 gives, say, a = 4, d = −1, i.e. the AP 4, 3, 2, 1, 0, −1, …. Here t5 = 0, confirming the (p + q) = 5th term is indeed 0. The reason students often miss it: the (p − q) factor cancels, and the surviving bracket is itself the (p + q)th term.

Final answer: 0