The ratio of sums of m and n terms of an A.P. is m² : n². Show that the ratio of with and nth term is 2m – 1 :2n – 1.

Let a be the first term and d be the common ratio.
    or      
or                         or      2an + n (m - 1)d = 2 am + m(n - 1)d
or           2am - 2an = (mn - n) d - (mn - m)d
or           2a (m - n) = d [mn -n -mn + m]
or           2a(m - n) = d (m -n)      or          d = 2a
          
∴     The ratio of mth term and nth term is:
                                  2m - 1 : 2n - 1