First, determine what type of sequence the set of numbers make up. Through simple logic, it is an arithmetic sequence, because one can see by inspection that there is a common difference of 3 (positive 3, just to be a bit more pedantic). We then use the formula, [latex] t_{n} = a + (n - 1)d[/latex] where [latex] t_{n} [/latex] represents the [latex] n^{th} [/latex] term; a represents the starting term (so the first number in the set of numbers, which in this case is -6); n is the term number (1st, 2nd, 3rd term, etc.); d is the common difference, that is, when you subtract the next term to the previous term – what is that numerical value. To elaborate a bit more, your 1st term is -6, 2nd is -3, 3rd is 0, etc. Also, the formula above is something you just learn, unless you learn to proof this formula, which is something different. So, here, [latex] t_{n} = -6 + (n-1)3[/latex], which can be expanded to: [latex] t_{n} = -6 + 3n-3[/latex] Therefore, [latex] t_{n} = 3n - 9[/latex]

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