#set document(title: "3.10 Arithmetic Sequences", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 3.10#h(0.6em)Arithmetic Sequences === Learning Objectives After completing this section, you should be able to: + Identify arithmetic sequences. + Find a given term in an arithmetic sequence. + Find the #math.equation(block: false, alt: "n")[$n$]th term of an arithmetic sequence. + Find the sum of a finite arithmetic sequence. + Use arithmetic sequences to solve real- world applications As we saw in the previous section, we are adding about 2.5 quintillion bytes of data per day to the Internet. If there are 550 quintillion bytes of data today, then there will be 552.5 quintillion bytes tomorrow, and 555 quintillion bytes in 2 days. This is an example of an arithmetic sequence. There are many situations where this concept of fixed increases comes into play, such as raises or table arrangements. === Identifying Arithmetic Sequences A #strong[sequence] of numbers is just that, a list of numbers in order. It can be a short list, such as the number of points earned on each assignment in a class, such as {10, 10, 8, 9, 10, 6, 10}. Or it can be a longer list, even infinitely long, such as the list of prime numbers. For example, here’s a sequence of numbers, specifically, the squares of the first 12 natural numbers. {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144} Each value in the sequence is called a #strong[term]. Terms in the list are often referred to by their location in the sequence, as in the #math.equation(block: false, alt: "n")[$n$]th term. For the sequence {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}, the first term of the sequence is 1, the fourth term is 16, and so on. In the sequence of assignment scores {10, 10, 8, 9, 10, 6, 10}, the first term is 10 and the third term is 8 . #figure(figph[A sequence of numbers. The numbers are as follows: 10, 10, 8, 9, 10, 6, and 10. A callout to the first 10 reads, First term, a sub 1. A callout to the second 10 reads, Second term, a sub 2. A callout to the third 10 reads, Fifth term, a sub 5.], alt: "A sequence of numbers. The numbers are as follows: 10, 10, 8, 9, 10, 6, and 10. A callout to the first 10 reads, First term, a sub 1. A callout to the second 10 reads, Second term, a sub 2. A callout to the third 10 reads, Fifth term, a sub 5.", caption: [Sequence showing first, second, and fifth terms]) The notation we use with sequences is a letter, which represents a term in the sequence, and a subscript, which indicates what place the term is in the sequence. For the sequence {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}, we will use the letter #math.equation(block: false, alt: "a")[$a$] as a value in the sequence, and so #math.equation(block: false, alt: "a sub 5")[$a_(5)$] would be the term in the sequence at the fifth position. That number is 25, so we can write #math.equation(block: false, alt: "a sub 5 equals 25")[$a_(5) = 25$]. In this section, we focus on a special kind of sequence, one referred to as an #strong[arithmetic sequence]. Arithmetic sequences have terms that increase by a fixed number or decrease by a fixed number, called the #strong[constant difference] (denoted by #math.equation(block: false, alt: "d")[$d$]), provided that value is not 0. This means the next term is always the previous term plus or minus a specified, constant value. Another way to say this is that the difference between any consecutive terms of the sequence is always the same value. To see a constant difference, look at the following sequence: {7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87}. illustrates that each term of the sequence is the previous term plus 8. Eight is the constant difference here. #figure(figph[A sequence of numbers. The numbers are as follows: 7, 15, 23, 31, 39, 47, 55, 63, 71, 79, and 87. Hops labeled plus 8 from each number points to the next number from left to right.], alt: "A sequence of numbers. The numbers are as follows: 7, 15, 23, 31, 39, 47, 55, 63, 71, 79, and 87. Hops labeled plus 8 from each number points to the next number from left to right.", caption: [Sequence of numbers with 8 added to each term]) #examplebox("Example 1")[Identifying Arithmetic Sequences][ Determine if the following sequences are arithmetic sequences. Explain your reasoning. + #math.equation(block: false, alt: "{ 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... }")[$\{ 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... \}$] + #math.equation(block: false, alt: "{ 20 , 40 , 80 , 160 , 320 , 640 }")[$\{ 20 , 40 , 80 , 160 , 320 , 640 \}$] + #math.equation(block: false, alt: "{ 7 , 1 , minus 5 , minus 11 , minus 17 , minus 23 , minus 29 , minus 34 , minus 40 }")[$\{ 7 , 1 , − 5 , − 11 , − 17 , − 23 , − 29 , − 34 , − 40 \}$] #solutionbox[ + In the sequence #math.equation(block: false, alt: "{ 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... }")[$\{ 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... \}$], every term is the previous term plus 3. The ellipsis indicates that the pattern continues, which means keep adding 3 to the previous term to get the new term. Therefore, this is an infinite arithmetic sequence. + In the sequence #math.equation(block: false, alt: "{ 20 , 40 , 80 , 160 , 320 , 640 }")[$\{ 20 , 40 , 80 , 160 , 320 , 640 \}$], terms increase by various amounts, for instance from term 1 to term 2, the sequence increases by 20, but from term 2 to term 3 the sequence increases by 40. So, this is not an arithmetic sequence. + In the sequence #math.equation(block: false, alt: "{ 7 , 1 , minus 5 , minus 11 , minus 17 , minus 23 , minus 29 , minus 34 , minus 40 }")[$\{ 7 , 1 , − 5 , − 11 , − 17 , − 23 , − 29 , − 34 , − 40 \}$], every term is the previous term minus 6, so this is an arithmetic sequence. ] ] Arithmetic sequences can be expressed with a formula. When we know the first term of an arithmetic sequence, which we label #math.equation(block: false, alt: "a sub 1")[$a_(1)$], and we know the constant difference, which is denoted #math.equation(block: false, alt: "d")[$d$], we can find any other term of the arithmetic sequence. The formula for the #math.equation(block: false, alt: "i th")[$i "th"$] term of an arithmetic sequence is #math.equation(block: false, alt: "a sub i equals a sub 1 plus d times open parenthesis i minus 1 close parenthesis")[$a_(i) = a_(1) + d × ( i − 1 )$]. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ If we have an arithmetic sequence with first term #math.equation(block: false, alt: "a sub 1")[$a_(1)$] and constant difference #math.equation(block: false, alt: "d")[$d$], then the #math.equation(block: false, alt: "i th")[$i "th"$] term of the arithmetic sequence is #math.equation(block: false, alt: "a sub i equals a sub 1 plus d times open parenthesis i minus 1 close parenthesis")[$a_(i) = a_(1) + d × ( i − 1 )$]. ] Let’s examine the formula with this arithmetic sequence: #math.equation(block: false, alt: "{ 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... }")[$\{ 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... \}$]. In this sequence #math.equation(block: false, alt: "a sub 1 equals 4")[$a_(1) = 4$] and #math.equation(block: false, alt: "d equals 3")[$d = 3$]. The table below shows the values calculated. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "i")[$i$], Place in Sequence], [#math.equation(block: false, alt: "a sub i")[$a_(i)$], #math.equation(block: false, alt: "i to the power t h")[$i^(t h)$] Term], [Value of Term], [Term Written as #math.equation(block: false, alt: "a sub 1 plus 3 times open parenthesis i minus 1 close parenthesis")[$a_(1) + 3 × ( i − 1 )$]]), [1], [#math.equation(block: false, alt: "a sub 1")[$a_(1)$]], [4], [#math.equation(block: false, alt: "4 plus 3 times 0")[$4 + 3 × 0$]], [2], [#math.equation(block: false, alt: "a sub 2")[$a_(2)$]], [7], [#math.equation(block: false, alt: "4 plus 3 times 1")[$4 + 3 × 1$]], [3], [#math.equation(block: false, alt: "a sub 3")[$a_(3)$]], [10], [#math.equation(block: false, alt: "4 plus 3 times 2")[$4 + 3 × 2$]], [4], [#math.equation(block: false, alt: "a sub 4")[$a_(4)$]], [13], [#math.equation(block: false, alt: "4 plus 3 times 3")[$4 + 3 × 3$]], [5], [#math.equation(block: false, alt: "a sub 5")[$a_(5)$]], [16], [#math.equation(block: false, alt: "4 plus 3 times 4")[$4 + 3 × 4$]], [#math.equation(block: false, alt: "i")[$i$]], [#math.equation(block: false, alt: "a sub i")[$a_(i)$]], [], [#math.equation(block: false, alt: "4 plus 3 times open parenthesis i minus 1 close parenthesis")[$4 + 3 × ( i − 1 )$]], )) We can see how the #math.equation(block: false, alt: "i th")[$i "th"$] term can be directly calculated. In this sequence, the formula is #math.equation(block: false, alt: "a sub 1 plus 3 times open parenthesis i minus 1 close parenthesis")[$a_(1) + 3 × ( i − 1 )$] where the first term, #math.equation(block: false, alt: "a sub 1")[$a_(1)$], is 4 and the constant difference #math.equation(block: false, alt: "d")[$d$] is 3. We can then determine the #math.equation(block: false, alt: "47 th")[$47 "th"$] term of this sequence: #math.equation(block: false, alt: "a sub 47 equals 4 plus 3 times open parenthesis 47 minus 1 close parenthesis equals 4 plus 3 times 46 equals 4 plus 138 equals 142")[$a_(47) = 4 + 3 × ( 47 − 1 ) = 4 + 3 × 46 = 4 + 138 = 142$]. #examplebox("Example 2")[Calculating a Term in an Arithmetic Sequence][ Identify #math.equation(block: false, alt: "a sub 1")[$a_(1)$] and #math.equation(block: false, alt: "d")[$d$] for the following arithmetic sequence. Use this information to determine the #math.equation(block: false, alt: "60 th")[$60 "th"$] term. #math.equation(block: true, alt: "{ 18 , 31 , 44 , 57 , 70 , 83 , ... }")[$\{ 18 , 31 , 44 , 57 , 70 , 83 , ... \}$] #solutionbox[ Inspecting the sequence shows that #math.equation(block: false, alt: "a sub 1 equals 18")[$a_(1) = 18$] and #math.equation(block: false, alt: "d equals 13")[$d = 13$]. We use those values in the formula, with #math.equation(block: false, alt: "i equals 60")[$i = 60$]. #math.equation(block: true, alt: "a sub 60 equals a sub 1 plus d times open parenthesis i minus 1 close parenthesis equals 18 plus 13 times open parenthesis 60 minus 1 close parenthesis equals 18 plus 13 times 59 equals 18 plus 767 equals 785")[$a_(60) = a_(1) + d × ( i − 1 ) = 18 + 13 × ( 60 − 1 ) = 18 + 13 × 59 = 18 + 767 = 785$] ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Arithmetic_Sequences")[Arithmetic Sequences] ] If we know two terms of the sequence, it is possible to determine the general form of an arithmetic sequence, #math.equation(block: false, alt: "a sub i equals a sub 1 plus d times open parenthesis i minus 1 close parenthesis")[$a_(i) = a_(1) + d × ( i − 1 )$]. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ If we have the #math.equation(block: false, alt: "i")[$i$]th term of an arithmetic sequence, #math.equation(block: false, alt: "a sub i")[$a_(i)$], and the #math.equation(block: false, alt: "j")[$j$]th term of the sequence, #math.equation(block: false, alt: "a sub j")[$a_(j)$], then the constant difference is #math.equation(block: false, alt: "d equals the fraction a sub j minus a sub i over j minus i")[$d = frac(a_(j) − a_(i), j − i)$] and the first term of the sequence is #math.equation(block: false, alt: "a sub 1 equals a sub i minus d open parenthesis i minus 1 close parenthesis")[$a_(1) = a_(i) − d ( i − 1 )$]. ] #examplebox("Example 3")[Determining First Term and Constant Difference Using Two Terms][ A sequence is known to be arithmetic. Two of its terms are #math.equation(block: false, alt: "a sub 7 equals 56")[$a_(7) = 56$] and #math.equation(block: false, alt: "a sub 19 equals 104")[$a_(19) = 104$]. Use that information to find the constant difference, the first term, and then the #math.equation(block: false, alt: "50 th")[$50 "th"$] term of the sequence. #solutionbox[ To find the constant difference, use #math.equation(block: false, alt: "d equals the fraction a sub j minus a sub i over j minus i")[$d = frac(a_(j) − a_(i), j − i)$]. The location of the terms is given by the subscript of the two #math.equation(block: false, alt: "a")[$a$] terms, #math.equation(block: false, alt: "i equals 7")[$i = 7$] and #math.equation(block: false, alt: "j equals 19")[$j = 19$]. So, the constant difference can be calculated as such: #math.equation(block: true, alt: "d equals the fraction 104 minus 56 over 19 minus 7 equals the fraction 48 over 12 equals 4")[$d = frac(104 − 56, 19 − 7) = frac(48, 12) = 4$] The constant difference of 4 is then used to find #math.equation(block: false, alt: "a sub 1")[$a_(1)$]. #math.equation(block: true, alt: "a sub 1 equals a sub i minus d open parenthesis i minus 1 close parenthesis equals a sub 7 minus 4 open parenthesis 7 minus 1 close parenthesis equals 56 minus 4 times 6 equals 32")[$a_(1) = a_(i) − d ( i − 1 ) = a_(7) − 4 ( 7 − 1 ) = 56 − 4 × 6 = 32$] So #math.equation(block: false, alt: "d equals 4")[$d = 4$] and #math.equation(block: false, alt: "a sub 1 equals 32")[$a_(1) = 32$]. With this information, the #math.equation(block: false, alt: "50 th")[$50 "th"$] term can be found. #math.equation(block: true, alt: "a sub 50 equals a sub 1 plus d times open parenthesis i minus 1 close parenthesis equals 32 plus 4 times open parenthesis 50 minus 1 close parenthesis equals 32 plus 4 times 49 equals 32 plus 196 equals 228")[$a_(50) = a_(1) + d × ( i − 1 ) = 32 + 4 × ( 50 − 1 ) = 32 + 4 × 49 = 32 + 196 = 228$] The #math.equation(block: false, alt: "50 th")[$50 "th"$] term is #math.equation(block: false, alt: "a sub 50 equals 228")[$a_(50) = 228$]. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Finding_the_First_Term_and_Constant_Difference_for_an_Arithmetic_Sequence")[Finding the First Term and Constant Difference for an Arithmetic Sequence] ] === Finding the Sum of a Finite Arithmetic Sequence Sometimes we want to determine the sum of the numbers of a finite arithmetic sequence. The formula for this is fairly straightforward. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ The sum of the first #math.equation(block: false, alt: "n")[$n$] terms of a finite arithmetic sequence, written #math.equation(block: false, alt: "s sub n")[$s_(n)$], with first and last term #math.equation(block: false, alt: "a sub 1")[$a_(1)$] and #math.equation(block: false, alt: "a sub n")[$a_(n)$], respectively, is #math.equation(block: false, alt: "s sub n equals n open parenthesis the fraction a sub 1 plus a sub n over 2 close parenthesis")[$s_(n) = n ( frac(a_(1) + a_(n), 2) )$]. ] #examplebox("Example 4")[Finding the Sum of a Finite Arithmetic Sequence][ What is the sum of the first 60 terms of an arithmetic sequence with #math.equation(block: false, alt: "a sub 1 equals 4.5")[$a_(1) = 4.5$] and #math.equation(block: false, alt: "d equals 2.5")[$d = 2.5$]? #solutionbox[ The formula requires the first and last terms of the sequence. The first term is given, #math.equation(block: false, alt: "a sub 1 equals 4.5")[$a_(1) = 4.5$]. The #math.equation(block: false, alt: "60 th")[$60 "th"$] term is needed. Using the formula #math.equation(block: false, alt: "a sub 1 equals a sub i plus d open parenthesis i minus 1 close parenthesis")[$a_(1) = a_(i) + d ( i − 1 )$] provides the value for the #math.equation(block: false, alt: "60 th")[$60 "th"$] term. #math.equation(block: true, alt: "a sub 60 equals 4.5 plus 2.5 open parenthesis 60 minus 1 close parenthesis equals 4.5 plus 2.5 times 59 equals 4.5 plus 147.5 equals 152")[$a_(60) = 4.5 + 2.5 ( 60 − 1 ) = 4.5 + 2.5 × 59 = 4.5 + 147.5 = 152$] Applying the formula #math.equation(block: false, alt: "s sub n equals n open parenthesis the fraction a sub 1 plus a sub n over 2 close parenthesis")[$s_(n) = n ( frac(a_(1) + a_(n), 2) )$] provides the sum of the first 60 terms. #math.equation(block: true, alt: "s sub 60 equals 60 open parenthesis the fraction 4.5 plus 152 over 2 close parenthesis equals 60 times the fraction 156.5 over 2 equals 4,695")[$s_(60) = 60 ( frac(4.5 + 152, 2) ) = 60 × frac(156.5, 2) = 4,695$] The sum of the first 60 terms is 4,695. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Finding_the_Sum_of_a_Finite_Arithmetic_Sequence")[Finding the Sum of a Finite Arithmetic Sequence] ] === Using Arithmetic Sequences to Solve Real-World Applications Applications of arithmetic sequences occur any time some quantity increases by a fixed amount at each step. For instance, suppose someone practices chess each week and increases the amount of time they study each week. The first week the person practices for 3 hours, and vows to practice 30 more minutes each week. Since the amount of time practicing increases by a fixed number each week, this would qualify as an arithmetic sequence. #examplebox("Example 5")[Applying an Arithmetic Sequence][ Jordan has just watched #emph[The Queen’s Gambit] and decided to hone their skills in chess. To really improve at the game, Jordan decides to practice for 3 hours the first week, and increase their time spent practicing by 30 minutes each week. How many hours will Jordan practice chess in week 20? #solutionbox[ Jordan’s practice scheme is an arithmetic sequence, as it increases by a fixed amount each week. The first week there are 3 hours of practice. This means #math.equation(block: false, alt: "a sub 1 equals 3")[$a_(1) = 3$]. Jordan increases the time spent practicing by 30 minutes, or half an hour, each week. This means #math.equation(block: false, alt: "d equals 0.5")[$d = 0.5$]. Using those values, and that we want to know the amount of time Jordan will study in week 20, we determine the time in week 20 using #math.equation(block: false, alt: "a sub i equals a sub 1 plus d times open parenthesis i minus 1 close parenthesis")[$a_(i) = a_(1) + d × ( i − 1 )$]. #math.equation(block: true, alt: "a sub 20 equals 3 plus 0.5 times open parenthesis 20 minus 1 close parenthesis equals 3 plus 0.5 times 19 equals 3 plus 9.5 equals 12.5")[$a_(20) = 3 + 0.5 × ( 20 − 1 ) = 3 + 0.5 × 19 = 3 + 9.5 = 12.5$] So, Jordan will practice 12.5 hours in week 20. ] ] #examplebox("Example 6")[Finding the Sum of a Finite Arithmetic Sequence][ Let’s check back in on Jordan. Recall, Jordan had just watched #emph[The Queen’s Gambit] and decided to hone their skills, practicing for 3 hours the first week, and increasing the time spent practicing by 30 minutes each week. How many hours total will Jordan have practiced chess after 30 weeks of practice? #solutionbox[ To calculate the total amount of time that Jordan practiced, we need to use #math.equation(block: false, alt: "s sub n equals n open parenthesis the fraction a sub 1 plus a sub n over 2 close parenthesis")[$s_(n) = n ( frac(a_(1) + a_(n), 2) )$]. The formula requires the first and last terms of the sequence. Since Jordan practiced 3 hours in the first week, the first term is #math.equation(block: false, alt: "a sub 1 equals 3")[$a_(1) = 3$]. Because we want the total practice time after 30 weeks, we need the #math.equation(block: false, alt: "30 th")[$30 "th"$] term. Because the constant difference is #math.equation(block: false, alt: "d equals 0.5")[$d = 0.5$], the #math.equation(block: false, alt: "30 th")[$30 "th"$] term is #math.equation(block: false, alt: "a sub 30 equals 3 plus 0.5 open parenthesis 30 minus 1 close parenthesis equals 3 plus 0.5 times 29 equals 3 plus 14.5 equals 17.5")[$a_(30) = 3 + 0.5 ( 30 − 1 ) = 3 + 0.5 × 29 = 3 + 14.5 = 17.5$]. Applying the formula #math.equation(block: false, alt: "s sub n equals n open parenthesis the fraction a sub 1 plus a sub n over 2 close parenthesis")[$s_(n) = n ( frac(a_(1) + a_(n), 2) )$] provides the sum of the first 30 terms. #math.equation(block: true, alt: "s sub 30 equals 30 open parenthesis the fraction 3 plus 17.5 over 2 close parenthesis equals 60 times the fraction 20.5 over 2 equals 615")[$s_(30) = 30 ( frac(3 + 17.5, 2) ) = 60 × frac(20.5, 2) = 615$] This means that Jordan practiced a total of 615 hours after 30 weeks. ] ] #notebox("Who Knew?", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[ #emph[The Fibonacci Sequence] Not all sequences are arithmetic. One special sequence is the #strong[Fibonacci sequence], which is the sequence that has as its first two terms 1 and 1. Every term thereafter is the sum of the previous two terms. The first nine terms of the Fibonacci sequence are 1, 1, 2, 3, 5, 8, 13, 21, and 34. This sequence is found in nature, architecture, and even music! In nature, the Fibonacci sequence describes the spirals of sunflower seeds, certain galaxy spirals, and flower petals. In music, the band Tool used the Fibonacci sequence in the song “Lateralus.” The Fibonacci sequence even relates to architecture, as it is closely related to the golden ratio. ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Fibonacci_Sequence_and_Lateralus")[Fibonacci Sequence and “Lateralus”] ] === Key Terms - sequence - term of a sequence - arithmetic sequence - first term - constant difference === Key Concepts - A sequence is a list of numbers. Any individual number in that list, or sequence, is a term of the sequence. A specific term of a sequence is denoted by the sequence symbol with a subscript indicating where the term in the sequence is. - A special form of a sequence is an arithmetic sequence. Each arithmetic sequence is determined by its first term and its constant difference. Any term in an arithmetic sequence is determined by adding the constant difference to the preceding term. - If the first term and the constant difference of an arithmetic sequence are known, then any term of the sequence can be found directly. - Because arithmetic sequences follow such a strict pattern, the sum of the first #math.equation(block: false, alt: "n")[$n$] terms of an arithmetic sequence can be determined with the formula #math.equation(block: false, alt: "s sub n equals n open parenthesis the fraction a sub 1 plus a sub n over 2 close parenthesis")[$s_(n) = n ( frac(a_(1) + a_(n), 2) )$]. === Formulas - #math.equation(block: false, alt: "a sub i equals a sub 1 plus d times open parenthesis i minus 1 close parenthesis")[$a_(i) = a_(1) + d × ( i − 1 )$] - #math.equation(block: false, alt: "d equals the fraction a sub j minus a sub i over j minus i")[$d = frac(a_(j) − a_(i), j − i)$] - #math.equation(block: false, alt: "a sub 1 equals a sub i minus d open parenthesis i minus 1 close parenthesis")[$a_(1) = a_(i) − d ( i − 1 )$] - #math.equation(block: false, alt: "s sub n equals n open parenthesis the fraction a sub 1 plus a sub n over 2 close parenthesis")[$s_(n) = n ( frac(a_(1) + a_(n), 2) )$] === Videos - #link("https://openstax.org/r/Arithmetic_Sequences")[Arithmetic Sequences] - #link("https://openstax.org/r/Finding_the_First_Term_and_Constant_Difference_for_an_Arithmetic_Sequence")[Finding the First Term and Constant Difference for an Arithmetic Sequence] - #link("https://openstax.org/r/Finding_the_Sum_of_a_Finite_Arithmetic_Sequence")[Finding the Sum of a Finite Arithmetic Sequence] - #link("https://openstax.org/r/Fibonacci_Sequence_and_Lateralus")[Fibonacci Sequence and “Lateralus”]