Q. 194.0( 2 Votes )

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

Answer :

Sum of m terms =

Sum of n terms =

Sum of m terms : Sum of n terms = m^{2} : n^{2}

⇒ : = m^{2} : n^{2}

⇒ n^{2}m(2a + (m–1)d) = nm^{2}( 2a + ( n–1)d)

⇒ 2an^{2}m + n^{2}m^{2}d – n^{2}md = 2anm^{2} + n^{2}m^{2}d – nm^{2}d

⇒ 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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