The ratio of the sums of m and n
Asked by admin @ in Math viewed by 338 People
the ratio of the sum of m and n terms of an AP is m2 :n2 show that the ratio of the mth and nth term is ( 2m-1):(2n-1)?
Answered by admin @
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]
m/2 [2a + (m -1)d] / n/2 [2a + (n -1)d] = m:n
⇒ [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 nth 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)
So, the ratio of mth term and the nth term of the arithmetic series is (2m - 1):(2n -1).
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