# Verify if the sequence n^2 + n + 1 is an a.p.Verify if the sequence n^2 + n + 1 is an a.p.

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an = n^2 + n + 1

Let us calculate some terms in the progression:

a1= 1^2 + 1 + 1 = 3

a2 = 2^2 + 2 +1 = 7

a3= 3^2 + 3 + 1 = 13

a4= 4^2 + 4 + 1 = 21

We notice that :

a2-a1 , a3-a2, and a4-a3 are not equals

Then the terms does not have any constant difference, then an is not an A.P.

We can establish if any series in which the nth term of the sequence is represented by f(n), is in A.P. by calculating the value of f(n+1) - f(n) and checking if it is a constant.

In the given sequence:

f(n) = n^2 + n + 1

and

f(n+1) = (n+1)^2 + (n +1) +1

= n^2 + 2n +1 + n +1 + 1

= n^2 + 3n + 3

Therefore:

f(n+1) - f(n) = (n^2 + 3n + 3) - (n^2 + n + 1)

= 2n + 2

This means that the difference between any two consecutive terms of the series is not constant. It is dependent on the value of n.

Therefore the series is not an A.P.

The sequence n^2+n+1 is an A.P. if for three consecutive values of n, the sum of the first and the third is twice that of the second.

Let n be the first value and we have the sequence n^2+n+1

Next take n+1, (n+1)^2+n+1+1=n^2+2n+1+n+1+1=n^2+3n+3

Next take n+2, (n+2)^2+n+2+1=n^2+4n+4+n+2+1=n^2+5n+7

Now adding the nth and (n+2)th terms we get 2*n^2+6n+8

Whereas 2 times the (n+1)th term is 2*n^2+6n+6

Therefore as 2*n^2+6n+6 is not equal to 2*n^2+6n+8, the given sequence is not an A.P.

To verify if the squence n^2+n+1 is in arith metic progression:

Let an = n^2+n+1

We can see that the successive terms has a difference:

an+1 - an = (n+1)^2+(n+1)+1 = n^2+n+1 = (n+1)^2-n^2+(n+1-n)

= 2n+1+1 = 2n+2.

So the difference between 1st and 2nd term = 1*2+2 =4.

When n=2, the difference beteen terms 2 and 3 = 2*2+2 =6.

The the difference is increasing from term to terms as n (term number ) increases. SO this is not an A P.

To verify if the given sequence is 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+n+1

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

a1 = 1^2 + 1 + 1

a1 = 3

a2 = 2^2 + 2 + 1

a2 = 7

a3 = 3^2 + 3 + 1

a3 = 13

a4 = 4^2 + 4 + 1

a4 = 21

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

a2 - a1 = 7 - 3 = 4

a3 - a2 = 13 - 7 = 6

a4 - a3 = 21 - 13 = 7

As we can see, the values obtained are not equal, they do not represent a common difference, so the given sequence is not an a.p.