If m times the mth term of an A. P. is equal to n times its nth term then show that the (m + n)th term of the A. P. is zero.

Solution: 

mt_m = nt_n  (Given)

∴ m [a + (m – 1) d] = n [a + (n – 1) d]      

                                        

                                                  [∵ t_n = a + (n – 1) d]

∴  m [a + md – d] = n [a + nd – d]

∴  am + m²d – md = an + n²d – nd

∴  am – an + m²d – n²d – md + nd = 0

∴  a (m – n) + d (m² – n²) – d (m – n) = 0

∴  a (m – n) + d (m + n) (m – n) – d (m – n) = 0    

                                                   

                                            [∵ a² – b² = (a + b) (a – b)]

Dividing through at by m – n, we get

a + d (m + n) - d = 0

∴  a + d [m + n – 1] = 0

∴  a + (m + n – 1) d = 0 ------------------------eq. no. (1)

t_m+n = a + (m + n – 1)   [∵ t_n = a + (n – 1) d]

∴ t_{m + n} = 0  [from eq. no. (1)]

Hence proved.