It is no more or n less significant than many other sequences.
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A geometric sequence is : a•r^n while a quadratic sequence is a• n^2 + b•n + c So the answer is no, unless we are talking about an infinite sequence of zeros which strictly speaking is both a geometric and a quadratic sequence.
You cant solve the next term (next number) in this sequence. You need more terms, because this is either a "quadratic sequence", or a "linear and quadratic sequence", and you need more terms than this to solve a "linear and quadratic sequence" and for this particular "quadratic sequence" you would need more terms to solve nth term, which would solve what the next number is. If this is homework, check with your teacher if he wrote the wrong sum.
These are called the second differences. If they are all the same (non-zero) then the original sequence is a quadratic.
It isn't clear what you want to justify.
To find the nth term of the sequence 5, 15, 29, 47, 69, we first determine the differences between consecutive terms: 10, 14, 18, and 22. The second differences are constant at 4, indicating that the nth term is a quadratic function. By fitting the quadratic formula ( an^2 + bn + c ) to the sequence, we find that the nth term is ( 2n^2 + 3n ). Thus, the nth term of the sequence is ( 2n^2 + 3n ).