Question

prove that the square of any term of the as 7,11,15,... will not be a term of the sequence​

Answer 1

Given AP is 7,11,15...

a = 7, d = 11 - 7 = 4

Square of any term is a^2.

Let (a + 4n)^2 = a^2

=> a^2 + 16n^2 + 8an = a^2

=> 16n^2 + 8an = 0

=> 16n^2 = 8an

=> 16n^2 = 8(7)(n)

=> 16n^2 = 56n

=> n = (16/56)

Since 15/56 is a rational number.

There are no natural numbers where there exists perfect square.

Answer 2

Hello siddharta7................