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...

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a ordered list of numbers in which each term after the first is obtained by adding a constant value to the preceding term. This constant is known as the common difference, and the sequence progresses linearly, making it predictable and easy to analyze mathematically.

Common Difference

The common difference in an arithmetic sequence is the fixed amount that each term increases or decreases by as you move from one term to the next. It is denoted by 'd' and defines the rate at which the sequence changes, allowing the computation of any term when combined with the first term.

General Term of an Arithmetic Sequence

The general term, or nth term, of an arithmetic sequence is given by the formula a? = a? + (n - 1)d, where a? is the first term and d is the common difference. This formula provides an explicit way to calculate any term in the sequence without having to list all the preceding terms.

Solving a System of Equations

When specific terms in an arithmetic sequence are given, such as the 7th and 17th terms, a system of linear equations can be set up using the general term formula. By solving this system, one can determine the values of the first term and the common difference. This approach applies basic algebraic techniques to find unknown parameters in sequences.

Transcript

00:01 So in this problem, we're given an arithmetic sequence, and we know the seventh term, a sub 7, is negative 1, and the 17th term, a sub 17, is negative 41.

00:10 So the first thing we need to do is find a sub 1 and d.

00:13 So to do this, we're going to use our a subn formula.

00:16 A sub n equals a sub 1 plus d times n minus 1.

00:20 And we're going to set up equation for both of the two terms we're given.

00:24 So if we start with a sub 7 equal to negative 1, remember, that means n is equal to 7.

00:30 So we would have negative 1 equal to a sub 1 plus d times 7 minus 1.

00:35 Well, 7 minus 1 is 6.

00:37 So we have negative 1 equal to a sub 1 plus 60.

00:41 Now we're going to do the same idea for our second term.

00:44 We know the 17th term is negative 41, so n is equal to 17.

00:48 So we would have negative 41 equal to a sub 1 plus d times 17 minus 1.

00:55 And 17 minus 1 is 16.

00:56 So we have negative 41 equal to a sub 1 plus 7.

01:00 16d.

01:01 Well, perfect.

01:02 Now we have a system of equations.

01:04 So i'm going to write our two equations together but flipped around, meaning i have a sub 1 plus 6d equal to negative 1, and a sub 1 plus 16d equal to negative 41.

01:16 And now i'm going to solve by elimination.

01:18 I'm going to subtract the two equations from each other because a sub 1 minus a sub 1 is 0...

Please give Ace some feedback