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## Algebra 1

### Course: Algebra 1 > Unit 9

Lesson 2: Constructing arithmetic sequences- Recursive formulas for arithmetic sequences
- Recursive formulas for arithmetic sequences
- Recursive formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Arithmetic sequence problem
- Converting recursive & explicit forms of arithmetic sequences
- Converting recursive & explicit forms of arithmetic sequences
- Converting recursive & explicit forms of arithmetic sequences
- Arithmetic sequences review

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# Explicit formulas for arithmetic sequences

CCSS.Math: ,

Learn how to find explicit formulas for arithmetic sequences. For example, find an explicit formula for 3, 5, 7,...

Before taking this lesson, make sure you are familiar with the basics of arithmetic sequence formulas.

## How explicit formulas work

Here is an explicit formula of the sequence 3, comma, 5, comma, 7, comma, point, point, point

In the formula, n is any term number and a, left parenthesis, n, right parenthesis is the n, start superscript, start text, t, h, end text, end superscript term.

This formula allows us to simply plug in the number of the term we are interested in, and we will get the value of that term.

In order to find the fifth term, for example, we need to plug n, equals, 5 into the explicit formula.

Cool! This is in fact the fifth term of 3, comma, 5, comma, 7, comma, point, point, point

### Check your understanding

## Writing explicit formulas

Consider the arithmetic sequence 5, comma, 8, comma, 11, comma, point, point, point The first term of the sequence is start color #0d923f, 5, end color #0d923f and the common difference is start color #ed5fa6, 3, end color #ed5fa6.

We can get any term in the sequence by taking the first term start color #0d923f, 5, end color #0d923f and adding the common difference start color #ed5fa6, 3, end color #ed5fa6 to it repeatedly. Check out, for example, the following calculations of the first few terms.

n | Calculation for the n, start superscript, start text, t, h, end text, end superscript term | ||
---|---|---|---|

1 | start color #0d923f, 5, end color #0d923f | equals, start color #0d923f, 5, end color #0d923f, plus, 0, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 5 | |

2 | start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, end color #ed5fa6 | equals, start color #0d923f, 5, end color #0d923f, plus, 1, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 8 | |

3 | start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, plus, 3, end color #ed5fa6 | equals, start color #0d923f, 5, end color #0d923f, plus, 2, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 11 | |

4 | start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, plus, 3, plus, 3, end color #ed5fa6 | equals, start color #0d923f, 5, end color #0d923f, plus, 3, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 14 | |

5 | start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, plus, 3, plus, 3, plus, 3, end color #ed5fa6 | equals, start color #0d923f, 5, end color #0d923f, plus, 4, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 17 |

The table shows that we can get the n, start superscript, start text, t, h, end text, end superscript term (where n is any term number) by taking the first term start color #0d923f, 5, end color #0d923f and adding the common difference start color #ed5fa6, 3, end color #ed5fa6 repeatedly for n, minus, 1 times. This can be written algebraically as start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis.

**In general, this is the standard explicit formula of an arithmetic sequence whose first term is start color #0d923f, A, end color #0d923f and common difference is start color #ed5fa6, B, end color #ed5fa6:**

### Check your understanding

## Equivalent explicit formulas

Explicit formulas can come in many forms.

For example, the following are all explicit formulas for the sequence 3, comma, 5, comma, 7, comma, point, point, point

- 3, plus, 2, left parenthesis, n, minus, 1, right parenthesis (this is the standard formula)

The formulas may look different, but the important thing is that we can plug an n-value and get the correct n, start superscript, start text, t, h, end text, end superscript term (try for yourselves that the other formulas are correct!).

Different explicit formulas that describe the same sequence are called

**equivalent formulas.**## A common misconception

An arithmetic sequence may have different equivalent formulas, but it's important to remember that

**only the standard form gives us the first term and the common difference.**For example, the sequence 2, comma, 8, comma, 14, comma, point, point, point has a first term of start color #0d923f, 2, end color #0d923f and a common difference of start color #ed5fa6, 6, end color #ed5fa6.

The explicit formula start color #0d923f, 2, end color #0d923f, start color #ed5fa6, plus, 6, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis describes this sequence, but the explicit formula start color #0d923f, 2, end color #0d923f, start color #ed5fa6, plus, 6, end color #ed5fa6, n describes a different sequence.

In order to bring the formula 2, plus, 6, left parenthesis, n, minus, 1, right parenthesis to an equivalent formula of the form A, plus, B, n, we can expand the parentheses and simplify:

Some people might prefer the formula minus, 4, plus, 6, n over the equivalent formula 2, plus, 6, left parenthesis, n, minus, 1, right parenthesis, because it's shorter. The nice thing about the longer formula is that it gives us the first term.

### Check your understanding

### Challenge problems

## Want to join the conversation?

- what dose it mean to create an explicit formula for a geometric(23 votes)
- An explicit formula directly calculates the term in the sequence that you want.

A recursive formula calculates each term based upon the value of the prior term. So, it usually takes more steps.(6 votes)

- What's the difference between this formula and a(n) = a(1) + (n - 1)d? Is one better or something?(7 votes)
- The reason we use a(n)= a+b( n-1 ), is because it is more logical in algebra(8 votes)

- Can you add a section on Simplifying Geometric and arithmetic equations? I have a test on this tomorrow, and there isn't a section to help me study... Wish me luck I guess :~:(10 votes)
- Determine the next 2 terms of this sequence

2,5,10,17.. then write the explicit form(1 vote)- warning: long answer

this isn't an arithmetic ("linear") sequence because the differences between the numbers are different (5-2=3, 10-5=5, 17-10=7)

however, you might notice that the differences of the differences between the numbers are equal (5-3=2, 7-5=2). that means the sequence is quadratic/power of 2.

since the sequence is quadratic, you only need 3 terms.

let x=the position of the term in the sequence

let y=the value of the term

the 1st term is 2, so x=1 and y=2

the 2nd term is 5, so x=2 and y=5

the 3rd term is 10, so x=3 and y=10

the function is y=ax^2+bx+c, so plug in each point to solve for a, b, and c.

(1,2): 2=a(1^2)+b(1)+c

(2,5): 5=a(2^2)+b(2)+c

(3,10): 10=a(3^2)+b(3)+c

simplify: 2=a+b+c

5=4a+2b+c

10=9a+3b+c

solve this using any method, but i'll use elimination:

10=9a+3b+c

-(5=4a+2b+c)

5=5a+b (equation 3 - equation 2)

5=4a+2b+c

-(2=a+b+c)

3=3a+b (equation 2 - equation 1)

then subtract the 2 equations just produced:

5=5a+b

-(3=3a+b)

2=2a

that means a=1.

substitute a=1 into 3=3a+b: 3=3+b, b=0.

substitute a and b into 2=a+b+c: 2=1+0+c, c=1

so the equation becomes y=1x^2+0x+1, or y=x^2+1

btw you can check (4,17) to make sure it's right(10 votes)

- how do you do this -3,-1/3,5/9,23/27,77/81,239/243(2 votes)
- The main thing to notice in your sequence is that there are actually 2 different patterns taking place --- one in the numerator and one in the denominator.

Looking at the denominators first, we have 1, 3 9, 27, 81, 243, ... (a geometric sequence).

Each successive term is multiplied by 3, so for any term n (where n>=1), its denominator would be 3^(n-1).

Then notice that every numerator is 4 less than its denominator, so the formula for the numerator would be 3^(n-1) - 4.

Putting these together, the terms of the sequence would be represented by the formula

( 3^(n-1) - 4 ) / 3^(n-1) where n>=1.

Hope this helps!(7 votes)

- To find the common difference between two terms, is taking the difference and dividing by the number of terms a viable workaround?

For example, the first term is 5 and the tenth term is 59. So take the difference, 59-5=54, then divide by the number of terms between them, which is 5. And the average difference would be 6. Does this always work?(2 votes)- Your shortcut is derived from the explicit formula for the arithmetic sequence like 5 + 2(n – 1) = a(n). Plug your numbers into the formula where x is the slope and you'll get the same result:

5 + x(10 – 1) = 59

5 + 9x = 59

9x = 54

x = 6

To find the slope, you take the difference between the 10th and the 1st term and divide it by the "# of additions" or by the difference between the 10th term and the 1st term.

So your shortcut works always, because it's the same thing as the explicit formula (5 + 6(10 – 1) = 59). Basically, it says to get the 10th number in the sequence, you have to start from the base of 5 then add 6 (the slope) to it not once, but in this case 9 times ("# of additions").

P.S. You have a typo in "divide by the number of terms between them, which is 5". It should be 9, not 5. Also the "average difference" should be called "common difference" or "slope".(6 votes)

- what is the recursive formula for airthmetic formula(4 votes)
- It seems to me that 'explicit formula' is just another term for iterative formulas, because both use the same form. Is this true? And is there another term for formulas using the
*m_ + _Bn*form as opposed to the*A_ + _B(n-1)*form or are they both referred to as explicit formulas?(3 votes)- m + Bn and A + B(n-1) are both equivalent explicit formulas for arithmetic sequences. A + B(n-1) is the standard form because it gives us two useful pieces of information without needing to manipulate the formula (the starting term A, and the common difference B).

An explicit formula isn't another name for an iterative formula. Even though they both find the same thing, they each work differently--they're NOT the same form.

In the iterative formula, "a(n-1)" means "the value of the (n-1)th term in the sequence", this is not "a times (n-1)."

In the explicit formula "d(n-1)" means "the common difference times (n-1), where n is the integer ID of term's location in the sequence."

Thankfully, you can convert an iterative formula to an explicit formula for arithmetic sequences. Converting is usually less work.**Take the iterative formula**:`a(1) = A`

`a(n) = a(n-1) + B`

(here a(n-1) is this function for the previous term, not multiplication)**Turn it into an explicit formula**by taking the initial term's value and adding it to B times the integer (n-1):`a(n) = A + B(n-1)`

(here B(n-1) is multiplication, not a function)(3 votes)

- How do you algebraically get

5+2(n-2) from

the standard form 3+2(n-1)?(1 vote)- 3 + 2(𝑛 − 1)

= [distribute the 2] = 3 + 2𝑛 − 2

= [add 2 and subtract 2] = 3 + 2 + 2𝑛 − 2 − 2 = 5 + 2𝑛 − 4

= [factor out 2] = 5 + 2(𝑛 − 2)(3 votes)

- There is a practice problem that I got for review for a test I have a few days that I don't really get. Basically it says:

*Her production follows the arithmetic sequence 299 + b, 222, 60b + 1, find:*

1) b

2) the first term

3) the common difference

4) the formula for the general term

So I don't really get what I am supposed to do with the given information.

Any help would be appreciated.(2 votes)