2A, The ratio of the sums of first m and first n terms of an A.P. is m

1.	2A, The ratio of the sums of first m and first n terms of an A.P. is m2 2. Show that thesratio ofmth and nth terms is (em : (2n1)
Answer» 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).

2A, The ratio of the sums of first m and first n terms of an A.P. is m2 2. Show that thesratio ofmth and nth terms is (em : (2n1)

Answer»

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).