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: 

mtm = nt(Given)

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

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

tm + n = 0  [from eq. no. (1)]

Hence proved.