Question

Prove that the square of any term of the sequence 1, 4, 7, 10.... also a term of it

Answer 1

Let us find out n'th term of this sequence which is an AP.

First term, a = 1

Common difference, d = 4 - 1 = 3

Then n'th term is,

→ a(n) = a + (n - 1)d

→ a(n) = 1 + (n - 1)3

→ a(n) = 1 + 3n - 3

→ a(n) = 3n - 2

This is the n'th term of the sequence.

Taking its square,

→ [a(n)]² = (3n - 2)²

→ [a(n)]² = 9n² - 12n + 4

→ [a(n)]² = 9n² - 12n + 6 - 2

→ [a(n)]² = 3(3n² - 4n + 2) - 2

Taking 3n² - 4n + 2 = n',

→ [a(n)]² = 3n' - 2

This means square of n'th term of the sequence is (n')'th term of the same sequence.

Hence Proved!