How do you find the nth term rule for 34,25,16,7,-2,...?

Oct 2, 2016

If you consider ${a}_{0} = 34$, then ${a}_{n} = {a}_{0} - 9 n$
If you consider ${a}_{1} = 34$, then ${a}_{n} = {a}_{0} - 9 \left(n - 1\right)$

Explanation:

First of all, we need to see that the rule of the sequence is "subtract nine", and I suppose that this is the easy part.

Now, we need to see the sequence with the point of view of the first term: we start with $34$, and obtain $25$ by subtracting $9$. Now, what's the link between the first and the third element? We obtain $16$ by subtracting $9$ again from $25$, which means that we are subtracting $9$ twice from $34$.

This should lead you to the general rule: start with a number, subtract $9$ once to obtain the second number, subtract $9$ twice to obtain the third number...and so on.

Using a proper notation, if we define ${a}_{0}$ as the first term of the sequence, we can write

${a}_{1} = {a}_{0} - 9$
${a}_{2} = {a}_{1} - 9 = \left({a}_{0} - 9\right) - 9 = {a}_{0} - 2 \cdot 9$
${a}_{3} = {a}_{2} - 9 = \left({a}_{0} - 2 \cdot 9\right) - 9 = {a}_{0} - 3 \cdot 9$
$\setminus \vdots$
${a}_{n} = {a}_{0} - n \cdot 9$

which would be your closed formula for ${a}_{n}$

NB: if you choose to start with ${a}_{1}$ instead of ${a}_{0}$, then you must rescale your coefficients, and you would have ${a}_{n} = {a}_{1} - \left(n - 1\right) \cdot 9$