The nth term of a sequence is 3n2-1 the nth term of a different sequence is 30-n2
find the only number in both sequences

The common number is 26.

Step-by-step explanation:

We know that the n-th term of a sequence is:

aₙ = 3*n^2 - 1

And the n-th term of another sequence is:

bₙ = 30 - n^2

Remember that in a sequence n is always an integer number.

We want to find a number that belongs to both sequences, then we want to find a pair of integers x and n, such that:

aₙ = bₓ

This is:

3*n^2 - 1 = 30 - x^2

Let's isolate one of the variables, i will isolate n.

3*n^2 = 30 - x^2 + 1 = 31 - x^2

n^2 = (31 - x^2)/3

n = √( (31 - x^2)/3)

Now we can try with different integer values of x, and see if n is also an integer.

if x = 1

n = √( (31 - 1^2)/3) = √10

We know that √10 is not an integer, so we need to try with another value of x.

if x = 2:

n = √( (31 - x^2)/3) = √(27/3) = √9 = 3

Then if we have x= 2, n is also an integer, n = 3.

Then we have:

a₃ = b₂

The common number between both sequences is:

a₃ = 3*(3)^2 - 1 = 26

b₂ = 30 - 2^2 = 26

answer: approximately 206 bones

i hope this u out a branlist and a would be greatly appreciated :

35 is an outlier

step-by-step explanation: