Question

If the ratio of sum of first 'm' and 'n' terms of anAP is m^2 : n^2,show that the ratio of its mth and nth terms of (2m-1) :(2n-1)

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Answer 1

Sm/Sn = m^2/n^2

Am/An = Sm-Sm-1/Sn-Sn-1

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

= m^2 - m^2 -1 +2m/ n^2 - n^2 -1 +2n

= 2m -1/ 2n -1

Answer 2

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^2 : n^2

= [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)

So, the ratio of m th term and the n th term of the arithmetic series is (2m - 1) : (2n -1).

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