The ratio of the sums of first m and first n terms of an A. P. is m2: n2. Show that the ratio of its mth and nth terms is (2m−1): (2n−1).

Question

Philoid · Accepted Answer

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

Sum of m terms of an A.P. = S_m = (m/2) × [2a + (m -1)d]
Sum of n terms of an A.P. = S_n = (n/2) × [2a + (n -1)d]
Given: S_m : S_n = m² : n²

⇒ 2an + mnd – nd = 2am + mnd – md

⇒ 2an – nd = 2am – md

⇒ 2an – 2am = nd – md

⇒ 2a(n - m) = d(n - m)

⇒ 2a = d

Now, Ratio of m^th term to n^th term:

Therefore, ratio of m^th term and the n^th term of the A.P. is

(2m - 1) : (2n -1).

The ratio of the sums of first m and first n terms of an A. P. is m²: n². Show that the ratio of its m^th and n^th terms is (2m−1): (2n−1).