If the sum of first p terms of an A.P. is a p 2 + b p , find its common difference.

Here, we are given, S p = a p 2 + b p Let us take the first term as a 1 and the common difference as d . Now, as we know, a p = S p - S p - 1 So, we get, a p = ( a p 2 + b p ) - [ a ( p - 1 ) 2 + b ( p - 1 ) ] = a p 2 + b p - [ a ( p 2 + 1 - 2 p ) + b p - b ] [ U s i n g ( a - b ) 2 = a 2 + b 2 - a b ] = a p 2 + b p - ( a p 2 + a - 2 a p + b p - b ) = a p 2 + b p - a p 2 - a + 2 a p - b p + b = 2 a p - a + b . . . . ( 1 ) Also, a p = a 1 + ( p - 1 ) d = a 1 + p d - d = p d + ( a 1 - d ) . . . . ( 2 ) On comparing the terms containing a p in ( 1 ) and ( 2 ) , we get, d p = 2 a p d = 2 a Therefore, the common difference is d = 2 a .

Write the n th term of the AP 1 m , 1 + m m , 1 + 2 m m , . . . .

Given AP is 1 m , 1 + m m , 1 + 2 m m , . . . . We know that the n th term of an A.P. is given by a n = a + ( n - 1 ) d . In the given AP, a = 1 m d = 1 + m m - 1 m = 1 + m - 1 m = 1 Substituting the values of a and d in the n th term formula. a n = 1 m + ( n - 1 ) 1 = 1 + ( n - 1 ) m m Thus, the n th term of the given AP is 1 + ( n - 1 ) m m .