Question

If p times the p^th term of an A.P. is equal to q times q^th term, then show that (p + q)^th term of that A.P. is zero (p ≠ q).

Answer 1

Let a be the first term and d be the common difference of the given A.P.

Then p times p^th term = q times q^th term    ...(Given)

∴ pt_p = qt_p

Now, t_p = a + (p − 1)d and t_q = a + (q − 1)d

∴ p[a + (p − 1)d] = q[a + (q − 1)d]

∴ p[a + (p − 1)d] − q[a + (q − 1)d] = 0

∴ ap + p(p − 1)d − qa − q(q − 1)d = 0

∴ ap − qa + (p − 1)d − q(q − 1)d = 0

∴ a(p − q) + (p² − p)d − (q² − q)d = 0

∴ a(p − q) + d[p² − p − q² + q] = 0

∴ a(p − q) + d[p² − q² − p + q] = 0

∴ a(p − q) + d[(p + q)(p − q) − (p − q)] = 0

∴ a(p − q) + d(p − q)(p + q − 1)] = 0

∴ (p − q) [a + d(p + q − 1)] = 0

∴ a + d(p + q − 1) = 0    ...[Dividing by (p − q). ∵ (p − q) ≠ 0]    ...(1)

Now, (p + q)th term of the A.P. = a + (p + q − 1)d     ...(2)

∴ from (1) and (2),

(p + q)^th term of the A.P. is zero.

i.e. `t_((p + q)) = 0`