Question

Prove that the arithmetic sequence 5, 8, 11 contains no perfect squares

Answer 1

Answer and Explanation:

Prove that the arithmetic sequence 5, 8, 11 contains no perfect squares ?

Solution :

The arithmetic sequence 5, 8, 11.

Here, First term a=5

The common difference d=8-5=3

The nth term of the sequence is

$a_n=a+(n-1)d$

$a_n=5+(n-1)3$

$a_n=5+3n-3$

$a_n=3n+2$

Let x be a natural number and its square is the nth term,

$x^2=3n+2$

$x^2-2=3n$

$n=\frac{x^2-2}{3}$

Now, for all integers from 0 to 9, n does not come out to be an integer.

Therefore, the arithmetic sequence 5, 8, 11, … contains no perfect squares.

Answer 2

Given :  arithmetic sequence 5, 8, 11

To find :  prove that arithmetic sequence does not contains perfect square

Solution:

a = 5

d = 3

aₙ = a  + (n - 1) d

= 5 + (n - 1) 3

= 5 + 3n - 3

= 3n + 2

Perfect square

k²  

k = 3m  or 3m + 1  or 3m+2

k = 3m

k²  = 9m²   = 3*3m²     = 3n  ≠  3n + 2

k = 3m + 1

k²   = (3m + 1)² = 9m² + 6m + 1

= 3m(3m +2)  + 1

= 3n  + 1  ≠  3n+ 2

k = 3m + 2

k²   = (3m + 2)² = 9m² + 12m + 4

= 9m² + 12m + 3 + 1

=  3(3m² + 4m + 1) + 1

= 3n  + 1  ≠  3n+ 2

Hence arithmetic sequence 5,8,11  does not contains any perfect square

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