Question

The ratio of the sums of first m and first n terms of an AP is m^2:n^2 . Show that the ratio of it's mth and nth terms is (2m-1) :(2n-1)
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Answer 1

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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