Question Video: Finding the Arithmetic Sequence Under a Certain Condition Mathematics

Video Transcript

Find the finite sequence 𝑎 sub 𝑛, given 𝑎 sub one is equal to negative 82, 𝑎 sub 12 is equal to negative 203, and the twelfth-to-last term is negative 115.

We begin by recalling that as our sequence is finite, it will have a fixed number of terms. We also recall that an arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant. This is called the common difference. For any arithmetic sequence, the formula for the 𝑛th term states that 𝑎 sub 𝑛 is equal to 𝑎 sub one plus 𝑛 minus one multiplied by 𝑑, where 𝑎 sub 𝑛 is the 𝑛th term, 𝑎 sub one is the first term, and 𝑑 is the common difference.

In this question, we are told that 𝑎 sub one, the first term, is negative 82. The twelfth term, 𝑎 sub 12, is equal to negative 203. Using the general formula, we see that 𝑎 sub 12 is equal to 𝑎 sub one plus 12 minus one multiplied by 𝑑. Substituting in the values of 𝑎 sub 12 and 𝑎 sub one gives us the equation negative 203 is equal to negative 82 plus 11𝑑. Adding 82 to both sides of this equation, we see that 11𝑑 is equal to negative 121. And dividing both sides of this equation by 11 gives us 𝑑 is equal to negative 11. The common difference of the finite sequence is negative 11.

As we now know the first term and the common deference, all that is left to find is the last term of the finite sequence. In order to do this, we’ll use the fact that the twelfth-to-last term is negative 115. If we let 𝑎 sub 𝑙 be the last term, then the sequence can be written as shown. This means that 𝑎 sub 𝑙 minus one will be the second-to-last term or the penultimate term. 𝑎 sub 𝑙 minus two will be the third-to-last term. We can continue this pattern to show that 𝑎 sub 𝑙 minus 11 will be the twelfth-to-last term in the sequence.

Using the general formula once again, this is equal to 𝑎 sub one plus 𝑙 minus 11 minus one multiplied by 𝑑. Substituting in our values for 𝑎 sub 𝑙 minus 11, 𝑎 sub one, and 𝑑 gives us the equation negative 115 is equal to negative 82 plus 𝑙 minus 12 multiplied by negative 11. We can add 82 to both sides of this equation such that negative 33 is equal to 𝑙 minus 12 multiplied by negative 11. Dividing both sides of this equation by negative 11, we have three is equal to 𝑙 minus 12. We can then add 12 to both sides of this equation such that 𝑙 is equal to 15.

We can therefore conclude that the sequence has 15 terms. This fifteenth or last term, 𝑎 sub 15, is equal to negative 82 plus 15 minus one multiplied by negative 11. 15 minus one is 14. And multiplying this by negative 11 gives us negative 154. Negative 82 minus 154 is equal to negative 236. After clearing some space, we can summarize what we have found. Our sequence has a common difference of negative 11, a first term of negative 82, and a last term of negative 236. The finite arithmetic sequence 𝑎 sub 𝑛 contains the terms negative 82, negative 93, negative 104, and so on, with a last term of negative 236.