# The ratio of the sums of first m and first n terms of an arithmetic series is m2: n2show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1)

Sum of m terms =

Sum of n terms =

Sum of m terms : Sum of n terms = m2 : n2

: = m2 : n2

n2m(2a + (m–1)d) = nm2( 2a + ( n–1)d)

2an2m + n2m2d – n2md = 2anm2 + n2m2d – nm2d

2anm(n–m) = nmd(n–m)

2a = d

mth term : nth term = a + (m–1)d : a + (n–1)d

mth term : nth term = a + (m–1)2a : a + (n–1) 2a

mth term : nth term = (2m–1) : (2n–1)

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