The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show tha

1.	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).
Answer» Solution: we have 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] = 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) So, the ratio of m th term and the n th term of the arithmetic series is (2m - 1) : (2n -1)

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).

Answer»

Solution: we have
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] = 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)

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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