For each arithmetic sequence described, find a1 and d and construct the sequence by stating the

For each arithmetic sequence described, find a1 and d and...

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. This fixed amount is known as the common difference, and the sequence is fully characterized by its first term and this common difference.

First Term

The first term, often denoted as a?, is the starting value of the arithmetic sequence. It is a crucial component because it, together with the common difference, determines every subsequent term in the sequence.

Common Difference

The common difference, represented by d, is the constant amount added to each term of the arithmetic sequence to yield the next term. It defines the rate at which the sequence increases or decreases.

General n-th Term Formula

The general formula for the n-th term of an arithmetic sequence is given by a? = a? + (n - 1)d. This formula allows one to compute any term in the sequence when the first term and the common difference are known, and it is essential for constructing and understanding the structure of the sequence.

Transcript

00:01 So in this problem, we have an arithmetic sequence, and we know that the ninth term, so a 9, a.

00:04 Sub 9, is equal to negative 19.

00:07 And the 21st term, a sub 21, is equal to negative 55.

00:11 So the first thing we need to do is find a sub 1 and d.

00:15 Well, in order to do this, we're going to use our a subn formula.

00:18 So a sub n is equal to a sub 1 plus d times m minus 1.

00:23 And we're going to set up a formula for each of these two scenarios.

00:26 So if a sub 9 is equal to negative 19, that means n is equal to 9 and a sub n is negative 19.

00:34 So we would have negative 19 equal to 9.

00:37 Actually, you know what? i'm going to put this a little bit off to the side here.

00:40 So we're going to have negative 19 equal to a sub 1 plus d times 9 minus 1.

00:46 And if i simplify this, we have 9 minus 1, which is 8.

00:49 So we have negative 19 equal to a sub 1 plus 8d.

00:54 Now we're going to do the same concept with the second term.

00:56 So because a sub 21 is equal to negative 55, we know that n is equal to 0 .5.

01:00 To 21 and a sub n is negative 55.

01:03 So we would have negative 55 equal to a sub 1 plus d times 21 minus 1.

01:10 Well, 21 minus 1 is 20.

01:12 So we have negative 55 equal to a sub 1 plus 20d.

01:16 So now we have a system of equations.

01:19 So i'm going to put them together.

01:21 And i'm actually going to switch the order, meaning i'll have a sub 1 plus 8d equal to negative 19, and then a sub 1 plus 20d is equal to negative 55.

01:33 So now i'm going to solve this using elimination.

01:35 So i'm going to go ahead and subtract the two equations from each other, because a sub 1 minus a sub 1 is zero.

01:41 So these cancel.

01:42 Then we have 8d minus 20d, which is negative 12d, and then negative 19 minus negative 55 is equal to 36...

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