Let a be the first and d be the common difference of an A.P. Since (m + 1)th term of an A.P. is twice the (n + 1)th term.

∴         
or    a + (m + 1 - 1)d = 2[a + (n + 1 - 1)d]
or          a + md = 2a + 2nd
or                       a = (m - 2n)d
Now,                
                                
                                                                       ...(i)
Now,           
                             
or                                                                    ...(ii)

From (i) and (ii), we have
                    

∴   (3m + 1)th term is twice the (m + n + 1)th term of an A.P.