Question

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

Answer 1

$\sf\frac{s_{m}}{s_{n}} = \frac{ \frac{ \cancel{m}}{ \cancel{2}} (2a + (m - 1)d)}{ \frac{ \cancel{n}}{ \cancel{2}}(2a + (n - 1)d)} = \frac{ {m}^{ \cancel{2}} }{ {n}^{ \cancel{2}} } \\ \\ \sf \frac{2a + md - d}{2a + nd - d} = \frac{m}{n} \\ \\ \sf 2am + \cancel{mnd} - dm = 2an + \cancel{mnd} - dn \\ \\ \sf 2am - 2an - dm + dn = 0 \\ \\ \sf 2a(m - n) - d(m - n) = 0 \\ \\ \sf (2a - d)(m - n) = 0 \\ \\ \sf 2a - d = 0 \\ \\ \sf d = 2a$

$\sf a_m = a + (m - 1)d \\ \\ \sf = a + (m - 1)2a \\ \\ = \sf a + 2am - 2a \\ \\ = \sf a(2m - 1)$

$\sf a_n = a + (n - 1)d \\ \\ \sf = a + (n - 1)2a \\ \\ = \sf a + 2an - 2a \\ \\ = \sf a(2n - 1)$

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

Hence, proved.