Question

If the ratio of the first m and n terms of an ap is m^2 and n^2 , show that the ratio of its mth and nth term is (2m - 1) ( 2n - 1)

Answer 1

Given that,

$\dfrac{S_m}{ S_n } = \dfrac{ {m}^{2} }{ {n}^{2} } \\ \\ \implies \dfrac{ \dfrac{ \cancel{m}}{ \cancel{2}}[2a + (m - 1)d] }{ \dfrac{ \cancel{n}}{ \cancel{2}}[2a + (n - 1)d] } = \dfrac{ {m}^{ \cancel{2}} }{ {n}^{ \cancel{2}} } \\ \\ \implies \dfrac{2a + md - d}{2a + nd - d} = \dfrac{m}{n} \\ \\ \implies 2an + { \cancel{mnd}} - dn = 2am + { \cancel{mnd}} - dm \\ \\ \implies 2an - 2am - dn + dm = 0 \\ \\ \implies 2a(n - m) - d(n - m) \\ \\ \implies (2a - d)(n - m) \\ \\ \implies 2a - d = 0 \\ \\ \implies 2a = d$



Now,

$a_m = a + (m - 1)d \\ \\ = a + (m - 1)2a \\ \\ = a + 2am - 2a \\ \\ = a(2m - 2 + 1) \\ \\ = a(2m - 1)...(i) \\ \\ \\ a_n = a + (n - 1)d \\ \\ = a + (n - 1)2a \\ \\ = a + 2an - 2a \\ \\ = a(2n - 2 + 1) \\ \\ = a(2n - 1)...(ii)$



Dividing (i) by (ii),

$\dfrac{a_m}{a_n} = \dfrac{{ \cancel{a}(2m - 1)}}{{ \cancel{a}}(2n - 1)} \\ \\ \dfrac{a_m}{a_n} = \dfrac{(2m - 1)}{(2n - 1)}$

Hence, proved.

Answer 2

Hope its help u...........