# How do you write the rule for the nth term given -5,10,-15,20,...?

Feb 20, 2017

n^("th")" term "->a_n=5n(-1)^(n+1)

#### Explanation:

Let the term count be $i$
Let the i'th term be ${a}_{i}$

Let the term the question askes for be ${a}_{n}$

Tip: Alternating positive an negative is achieved using something like: ${\left(- 1\right)}^{n}$. So some variant on this will be included.

$\textcolor{b r o w n}{\text{Just considering the numbers. Will deal with the signs afterwards.}}$

${a}_{1} \to 5$
${a}_{2} \to 10 \to 10 - 5 = 5$
${a}_{3} \to 15 \to 15 - 10 = 5$
${a}_{4} \to 20 \to 20 - 15 = 5$

So this is an arithmetic sequence of type ${a}_{i} = 5 i$
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$\textcolor{b r o w n}{\text{Now we deal with the sign}}$

${a}_{1} > 0 \leftarrow \text{ } {\left(- 1\right)}^{2}$
${a}_{2} < 0 \leftarrow \text{ } {\left(- 1\right)}^{3}$
${a}_{3} > 0 \leftarrow \text{ } {\left(- 1\right)}^{4}$

So for any $i \text{ ; "a_i" includes } {\left(- 1\right)}^{i + 1}$

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$\textcolor{b r o w n}{\text{Putting it all together}}$

For any $i \text{; } {a}_{i} = 5 i {\left(- 1\right)}^{i + 1}$

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$\textcolor{b r o w n}{\text{Putting it into the same format the question requires}}$

n^("th")" term "->a_n=5n(-1)^(n+1)