Question

The ratio of sum of first m and n terms of an ap is m^2/n^2. prove that the ratio of mth and nth term is (2m-1)/(2n-1)

Answer 1

Given :
Sum of m terms of an A.P. = m/2 [2a + (m -1)d]

Sum of n terms of an A.P. = n/2 [2a + (n -1)d]

Solution:

m/2 [2a + (m -1)d] / n/2 [2a + (n -1)d] = m2 : n2
⇒ [2a + md - d] / [2a + nd - d] = m/n
⇒ 2an + mnd - nd = 2am + mnd - md
⇒ 2an - 2am = nd - md
⇒ 2a (n -m) = d(n - m)
⇒ 2a = d
Ratio of m th term to n th term:
[a + (m - 1)d] / [a + (n - 1)d]
= [a + (m - 1)2a] / [a + (n - 1)2a]
= a [1 + 2m - 2] / a[1 + 2n -2]
= (2m - 1) / (2n -1)
Hence,Proved.

This is the answer.
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