Question

The ratio of the sums of first m and first n terms of an arithmetic series is m² : n² show that the ratio of the m^th and n^th terms is (2m – 1) : (2n - 1).

Answer 1

Given,

n[2a + (m – 1)d] = m[2a + nd - d] 

2 an + mnd - nd = 2 am + mnd — md 

2an - 2am = nd - md 

2 a(n - m) = d(n - m) 

÷ by (n – m) on both sides, 

2a = d 

To prove, t_m : t_n = (2m – 1) : (2n – 1) 

L.H.S = t_m : t_n

= a + (m – 1) d : a + (n – 1)d 

= a + (m – 1) 2a : a + (n – 1)2a 

[Substitute the value of d = 2a]

= a + 2 am – 2 a: a + 2 am – 2a 

= 2am – a : 2an – a 

= a (2m – 1) : a (2n – 1) 

= (2m – 1) : (2n – 1) = R. H. S 

= (2m – 1) : (2n – 1) 

∴ t_m : t_n

L.H.S = R.H.S 

Hence it is proved.