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

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The nth term of an AP is a + (n - 1)d, where a is the first term and d is the common difference. The sum of n terms of an AP is (2a + (n - 1)d)(n/2)

Here Sm/Sn = [(2a + (m - 1)d)(m/2)]/[(2a + (n - 1)d)(n/2)] = m^2/n^2

=> [(2a + (m - 1)d)]/[(2a + (n - 1)d)] = m/n

=> n(2a + (m - 1)d) = m(2a + (n - 1)d)

=> 2an + n(m - 1)d = 2am + m(n -1)d

=> 2an + nmd - nd = 2am + mnd - md

=> 2an - nd = 2am - md

=> n(2a - d) = m(2a - d)

=> m = n

The ratio of the mth and nth terms is (a + (m - 1)d)/(a + (n - 1)d)

=> (a + (n - 1)d)/(a + (n - 1)d)

=> 1

=> (2m - 1)/(2n - 1)

**As m = n, we prove that Tm/Tn = (2m - 1)/(2n - 1)**