Question

How do you write a recursive rule and an explicit rule for the arithmetic sequence $19,9,-1,-11,.....$?

Answer 1

Hint: In this question we have been given with a sequence of numbers which are decreasing by a common factor of $10$. This implies that the difference between any $2$ consecutive terms will be $10$. We will make a rule which will tell us the value of the ${{n}^{th}}$ term in the sequence. We will consider the value of the ${{n}^{th}}$ term to be ${{a}_{n}}$ and get the required solution.

Complete step by step answer:
We have the sequence given to us as $19,9,-1,-11,.....$
This implies that the series is recurring. We have to find a rule which will give us the value of the ${{n}^{th}}$ term in the sequence.
We can see that the common difference is $10$. This implies that if we know the ${{n}^{th}}$ term, we can get the ${{\left( n+1 \right)}^{th}}$ term by subtracting $10$ from it.
We know the value of the first term in the sequence as:
$\Rightarrow {{a}_{0}}=19$
Now the next terms value can be calculated as ${{a}_{1}}=19-10$, which gives us the value ${{a}_{1}}=9$ and so on for the value of ${{a}_{n}}$.
So, we can conclude that the rule applies for us to start from the initial value $19$, and subtract $10$ with each iteration.
This means that, after $n$ iterations, we will have subtracted ten $n$ times, that is, we will have to subtract a total of $10n$.
Therefore, the rule for our sequence will be:
$\Rightarrow {{a}_{n}}=19-10n$, which is the required solution.

Note: In this question we were given with a sequence of numbers in arithmetic progression with the common difference between the terms as $-10$. The formula for the ${{n}^{th}}$ term in an arithmetic progression should be remembered which is ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$, where ${{a}_{n}}$ is the value of the ${{n}^{th}}$ term, and $d$ is the common difference.