# Prove that the sequence n^2+4 is not an A.P.

crmhaske | Certified Educator

The simple answer is it is not an A.P. because A.P.'s do not contain powers; however, to prove it:

a1 = 1 + 4 = 5
a2 = 4 + 4 = 8
a3 = 9 + 4 = 13
a4 = 16 + 4 = 20

a2 - a1 = 3
a3 - a2 = 5
a4 - a3 = 7

As the differences between terms are not equal, this is not an arithmetic series.

hala718 | Certified Educator

n^2 + 4

Let us substitute:

a1= 1 + 4 = 5

a2= 2^2 + 4 = 8

A3= 3^2 + 4 = 13

a4 = 4^2 + 4 = 20

If the terms are part of an arthimatical progression:

==> a2-a1 = a3-a2 = a4-a3

==> 8-5 = 13-8 = 20 - 13

==> 3  = 5 = 7

Then it is NOT an A.P

giorgiana1976 | Student

To demonstrate that the given sequence is not an a.p., we'll have to calculate the differences between 2 consecutive terms of the sequence and to prove that the differences obtained are not equal.

Let's put an = n^2+4

Now, we'll calculate the first 4 terms of the sequence:

a1 = 1^2 + 4

a1 = 5

a2 = 2^2 + 4

a2 = 8

a3 = 3^2 + 4

a3 = 13

a4 = 4^2 + 4

a4 = 20

Now, we'll verify if the values of the differences a2 - a1,  a3 - a2 and a4 - a3, are equal.

a2 - a1 = 8-5 = 3

a3 - a2 = 13 - 8 = 5

a4 - a3 = 20 - 13 = 7

As we can see, the values obtained are not equal, so the given sequence is not an a.p.

neela | Student

The nth term  of the sequence Tn = n^2+4.

Tr+1  = (r+1)^2+4

Tr = r^2

Therefore Tr+1 - Tr = (r+1)^2+4-r^2-4 = 2r^2+2r+1-r^2 = 2r+1.

Therefore the difference between the successive terms = 2r+1 is not a constant .

T2-T1 =  2^2+4 -1^2-4 = 3

T3-T2 =3^2+4-2^2-4 = 9-4 =5.

Thus the diffrence between the succesive terms is not a constant. So the sequence is not in AP.