#set document(title: "9.2 Arithmetic and Geometric 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")) == 9.2#h(0.6em)Arithmetic and Geometric Sequences === Arithmetic Sequences A charter tour bus service charges \$50 plus \$15 for each passenger. The cost of a tour is then a function of the number of passengers. Because the number of passengers, #math.equation(block: false, alt: "n")[$n$], can only be a positive integer, the function is a sequence. If #math.equation(block: false, alt: "C sub n")[$C_(n)$] represents the cost of a tour for passengers, then #math.equation(block: true, alt: "C sub 1, equals 50 plus 15 open parenthesis 1 close parenthesis equals 65; C sub 2, equals 50 plus 15 open parenthesis 2 close parenthesis equals 80; C sub 3, equals 50 plus 15 open parenthesis 3 close parenthesis equals 95")[$C_(1) & = 50 + 15 ( 1 ) = 65 \ C_(2) & = 50 + 15 ( 2 ) = 80 \ C_(3) & = 50 + 15 ( 3 ) = 95$] and in general #math.equation(block: true, alt: "C sub n equals 50 plus 15 n")[$C_(n) = 50 + 15 n$] Each term of this sequence can be obtained from the previous one by adding 15. A sequence in which each term can be obtained from the previous term by adding a fixed amount is called an #strong[arithmetic sequence]. The fixed amount that we add to each term is the difference between two successive terms and is called the #strong[common difference]. In the example above the common difference is 15. If we denote the first term of an arithmetic sequence by #math.equation(block: false, alt: "a")[$a$] and the common difference by #math.equation(block: false, alt: "d")[$d$], then the sequence can be defined recursively by #math.equation(block: true, alt: "a sub 1, equals a; a sub n plus 1, equals a sub n plus d")[$a_(1) & = a \ a_(n + 1) & = a_(n) + d$] #examplebox("Example 1")[][ Find the first four terms of an arithmetic sequence with first term 6 and common difference 3. #solutionbox[ The first term is 6, so we have #math.equation(block: false, alt: "a sub 1 equals 6")[$a_(1) = 6$]. To find each subsequent term, we add 3 to the previous term: #math.equation(block: true, alt: "a sub 2, equals a sub 1 plus 3 equals 6 plus 3 equals 9; a sub 3, equals a sub 2 plus 3 equals 9 plus 3 equals 12; a sub 4, equals a sub 3 plus 3 equals 12 plus 3 equals 15")[$a_(2) & = a_(1) + 3 = 6 + 3 = 9 \ a_(3) & = a_(2) + 3 = 9 + 3 = 12 \ a_(4) & = a_(3) + 3 = 12 + 3 = 15$] The first four terms are #math.equation(block: false, alt: "6 ,")[$6 ,$] #math.equation(block: false, alt: "9 ,")[$9 ,$] #math.equation(block: false, alt: "12 ,")[$12 ,$] and #math.equation(block: false, alt: "15")[$15$]. ] ] Find the first four terms of an arithmetic sequence with first term #math.equation(block: false, alt: "100")[$100$] and common difference #math.equation(block: false, alt: "minus 30")[$− 30$]. Answer: \_\_\_\_\_ (List the terms in order; separate terms with a comma. #solutionbox[ 100, 70, 40, 10 ] Find the first four terms of an arithmetic sequence with first term #math.equation(block: false, alt: "100")[$100$] and common difference #math.equation(block: false, alt: "minus 30")[$− 30$]. #solutionbox[ 100, 70, 40, 10 ] An arithmetic sequence defines a linear function of #math.equation(block: false, alt: "n")[$n$]. In the figure below, compare the graph of the linear function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x plus 3")[$f ( x ) = 2 x + 3$], whose domain is the set of all real numbers, and the graph of the arithmetic sequence #math.equation(block: false, alt: "a sub n equals 2 n plus 3")[$a_(n) = 2 n + 3$], whose domain is the set of positive integers. #figure(figph[line 2x plus 3], alt: "line 2x plus 3", caption: none) #figure(figph[sequence 2n plus 3], alt: "sequence 2n plus 3", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Notice that the common difference, 2, of the sequence corresponds to the slope of the linear function. ] Fill in the blanks. + To find the next term of an arithmetic sequence, we \_\_\_\_\_ a fixed amount. + The fixed amount is called the \_\_\_\_\_. + An arithmetic sequence defines \_\_\_\_\_ function of #math.equation(block: false, alt: "n")[$n$]. #solutionbox[ + add + common difference + a linear ] Fill in the blanks. + To find the next term of an arithmetic sequence, we \_\_\_\_\_\_\_\_\_\_ a fixed amount. + The fixed amount is called the \_\_\_\_\_\_\_\_\_\_. + An arithmetic sequence defines \_\_\_\_\_\_\_\_\_\_ function of #math.equation(block: false, alt: "n")[$n$]. === The General Term of an Arithmetic Sequence We have found a recursive definition for an arithmetic sequence, but we can also find a non-recursive definition. That is, we can find a formula for the general term. Consider an arithmetic sequence with first term #math.equation(block: false, alt: "a")[$a$] and common difference #math.equation(block: false, alt: "d")[$d$]. The #math.equation(block: true, alt: "first term is, a; second term is, a plus d; third term is, a plus d plus d equals a plus 2 d; fourth term is, a plus d plus d plus d equals a plus 3 d; n to the power th term is, a plus d plus d plus ⋯ plus d equals a plus open parenthesis n minus 1 close parenthesis d")[$"first term is" " " " " " " " " " " " " & a \ "second term is" " " " " " " " " " " " " & a + d \ "third term is" " " " " " " " " " " " " & a + d + d = a + 2 d \ "fourth term is" " " " " " " " " " " " " & a + d + d + d = a + 3 d \ n^("th") " " "term is" " " " " " " " " " " " " & a + d + d + ⋯ + d = a + ( n − 1 ) d$] Thus, we have the following property. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Arithmetic Sequence] The #math.equation(block: false, alt: "n to the power th")[$n^("th")$] term of an arithmetic sequence is #math.equation(block: true, alt: "a sub n equals a plus open parenthesis n minus 1 close parenthesis d")[$a_(n) = a + ( n − 1 ) d$] ] We can use this formula to find a particular term of an arithmetic sequence if we know the first term and the common difference. #examplebox("Example 2")[][ Find the fourteenth term of the arithmetic sequence #math.equation(block: false, alt: "minus 6 , minus 1 , 4 , ⋯")[$− 6 , " " − 1 , " " 4 , " " ⋯$] #solutionbox[ First, we find the common difference by subtracting any term from its successor: #math.equation(block: true, alt: "d equals minus 1 minus open parenthesis minus 6 close parenthesis equals 5")[$d = − 1 − ( − 6 ) = 5$] Then we use the formula with #math.equation(block: false, alt: "n equals 14")[$n = 14$] : #math.equation(block: true, alt: "a sub 14 equals minus 6 plus open parenthesis 14 minus 1 close parenthesis open parenthesis 5 close parenthesis equals 59")[$a_(14) = − 6 + ( 14 − 1 ) ( 5 ) = 59$] ] ] Find the 100th term of the arithmetic sequence #math.equation(block: false, alt: "1 , 3 , 5 , 7 , ⋯")[$1 , 3 , 5 , 7 , ⋯$]. Answer: \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "199")[$199$] ] Find the 100th term of the arithmetic sequence #math.equation(block: false, alt: "1 , 3 , 5 , 7 , ⋯")[$1 , 3 , 5 , 7 , ⋯$]. #solutionbox[ #math.equation(block: true, alt: "199")[$199$] ] #examplebox("Example 3")[][ You deposit \$8000 in an account that pays 5% simple annual interest. + Find a formula for the amount of money in the account after #math.equation(block: false, alt: "n")[$n$] years. + How much money will be in the account after 20 years? #solutionbox[ + Each year, 0.05(8000) or \$400 interest is added to the account balance. Thus, the annual balances, #math.equation(block: false, alt: "b sub n")[$b_(n)$], form an arithmetic sequence whose first term is 8400 and whose common difference is 400. Hence, #math.equation(block: true, alt: "b sub n equals 8400 plus open parenthesis n minus 1 close parenthesis 400")[$b_(n) = 8400 + ( n − 1 ) 400$] + We substitute #math.equation(block: false, alt: "n equals 20")[$n = 20$] into the formula for the general term found in part (a). #math.equation(block: true, alt: "b sub 20 equals 8400 plus open parenthesis 19 close parenthesis 400 equals 16 , 000")[$b_(20) = 8400 + ( 19 ) 400 = 16 , 000$] After 20 years there will be \$16,000 in the account. ] ] Find the #math.equation(block: false, alt: "n")[$"n"$]th term of an arithmetic sequence whose first term is #math.equation(block: false, alt: "a")[$"a"$] and whose common difference is #math.equation(block: false, alt: "b")[$"b"$]. Answer: \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "open bracket @ $ a @ close bracket plus open bracket @ $ b @ close bracket open parenthesis open bracket @ $ n @ close bracket minus 1 close parenthesis equals open bracket @ $ t e r m @ close bracket")[$[ @ \$ a @ ] + [ @ \$ b @ ] ( [ @ \$ n @ ] − 1 ) = [ @ \$ t e r m @ ]$] ] Find the #math.equation(block: false, alt: "n")[$n$]th term of an arithmetic sequence whose first term is #math.equation(block: false, alt: "1000")[$1000$] and whose common difference is #math.equation(block: false, alt: "15")[$15$]. #solutionbox[ #math.equation(block: true, alt: "1000 plus 15 open parenthesis n minus 1 close parenthesis")[$1000 + 15 ( n − 1 )$] ] === Geometric Sequences A national junior chess tournament starts with 1024 invited contestants. At the end of each round, the winners move on to the next level. Thus, after each round there are half as many contestants as before, and the number of remaining contestants is a function of the number of rounds completed. Because the number of rounds #math.equation(block: false, alt: "n")[$n$] is a positive integer, the function is a sequence. If #math.equation(block: false, alt: "C sub n")[$C_(n)$] represents the number of contestants after #math.equation(block: false, alt: "n")[$n$] rounds of competition, then #math.equation(block: true, alt: "C sub 1, equals 1024 open parenthesis the fraction 1 over 2 close parenthesis equals 512; C sub 2, equals 512 open parenthesis the fraction 1 over 2 close parenthesis equals 1024 open parenthesis the fraction 1 over 2 close parenthesis squared equals 256; C sub 3, equals 256 open parenthesis the fraction 1 over 2 close parenthesis equals 1024 open parenthesis the fraction 1 over 2 close parenthesis cubed equals 128")[$C_(1) & = 1024 ( display(frac(1, 2)) ) = 512 \ C_(2) & = 512 ( display(frac(1, 2)) ) = 1024 attach(( display(frac(1, 2)) ), t: 2) = 256 \ C_(3) & = 256 ( display(frac(1, 2)) ) = 1024 attach(( display(frac(1, 2)) ), t: 3) = 128 " "$] In general #math.equation(block: true, alt: "C sub n equals 1024 open parenthesis the fraction 1 over 2 close parenthesis to the power n")[$C_(n) = 1024 attach(( display(frac(1, 2)) ), t: n)$] Each term of the sequence defined above can be obtained from the previous one by multiplying by #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]. A sequence in which each term can be obtained from the previous term by multiplying by a fixed amount is called a #strong[geometric sequence]. The fixed amount we multiply each term by is the ratio of two successive terms and is called the #strong[common ratio]. In the example above the common ratio is #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]. If we denote the first term of a geometric sequence by #math.equation(block: false, alt: "a")[$a$] and the common ratio by #math.equation(block: false, alt: "r")[$r$], then the sequence can be defined recursively by #math.equation(block: true, alt: "a sub 1, equals a; a sub n plus 1, equals r a sub n")[$a_(1) & = a \ a_(n + 1) & = r a_(n)$] #examplebox("Example 4")[][ Find the first four terms of a geometric sequence whose first term is 64 and whose common ratio is #math.equation(block: false, alt: "the fraction 5 over 4")[$display(frac(5, 4))$]. #solutionbox[ The first term is 64, so we have #math.equation(block: false, alt: "a sub 1 equals 64")[$a_(1) = 64$]. To find each subsequent term, we multipy the previous term by #math.equation(block: false, alt: "the fraction 5 over 4")[$display(frac(5, 4))$]. #math.equation(block: true, alt: "a sub 2, equals open parenthesis the fraction 5 over 4 close parenthesis a sub 1 equals open parenthesis the fraction 5 over 4 close parenthesis open parenthesis 64 close parenthesis equals 80; a sub 3, equals open parenthesis the fraction 5 over 4 close parenthesis a sub 2 equals open parenthesis the fraction 5 over 4 close parenthesis open parenthesis 80 close parenthesis equals 100; a sub 4, equals open parenthesis the fraction 5 over 4 close parenthesis a sub 3 equals open parenthesis the fraction 5 over 4 close parenthesis open parenthesis 100 close parenthesis equals 125")[$a_(2) & = ( display(frac(5, 4)) ) a_(1) = ( display(frac(5, 4)) ) ( 64 ) = 80 \ a_(3) & = ( display(frac(5, 4)) ) a_(2) = ( display(frac(5, 4)) ) ( 80 ) = 100 \ a_(4) & = ( display(frac(5, 4)) ) a_(3) = ( display(frac(5, 4)) ) ( 100 ) = 125$] The first four terms are 64, 80, 100, and 125. ] ] Find the first four terms of a geometric sequence whose first term is 5 and whose common ratio is 2. Answer: \_\_\_\_\_ (List the terms in order; separate terms with a comma. #solutionbox[ 5, 10, 20, 40 ] Find the first four terms of a geometric sequence whose first term is 5 and whose common ratio is 2. #solutionbox[ 5, 10, 20, 40 ] A geometric sequence defines an exponential function of #math.equation(block: false, alt: "n")[$n$]. In the figure below, compare the graphs of the exponential function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 100 open parenthesis 2 close parenthesis to the power x minus 1")[$f ( x ) = 100 ( 2 )^(x − 1)$], whose domain is the set of real numbers, and the geometric sequence #math.equation(block: false, alt: "a sub n equals 100 open parenthesis 2 close parenthesis to the power n minus 1")[$a_(n) = 100 ( 2 )^(n − 1)$], whose domain is the set of positive integers. #figure(figph[exponential function], alt: "exponential function", caption: none) #figure(figph[geometric sequence], alt: "geometric sequence", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Notice that the common ratio of the geometric sequence corresponds to the base of the exponential function. ] Recall that an arithmetic sequence defines a linear function whose slope corresponds to the common difference. Just as an exponential function grows much faster in the long run than a linear function, so does a geometric sequence grow much faster than an arithmetic sequence. Fill in the blanks. + To find the next term of a geometric sequence, we \_\_\_\_\_ a fixed amount. + The fixed amount is called the \_\_\_\_\_. + A geometric sequence defines \_\_\_\_\_ function of #math.equation(block: false, alt: "n")[$n$]. #solutionbox[ + multiply by + common ratio + an exponential ] Fill in the blanks. + To find the next term of a geometric sequence, we \_\_\_\_\_\_\_\_\_\_ + The fixed amount is called the \_\_\_\_\_\_\_\_\_\_. + A geometric sequence defines \_\_\_\_\_\_\_\_\_\_ function of #math.equation(block: false, alt: "n")[$n$]. #examplebox("Example 5")[][ Identify the following sequences as arithmetic, geometric, or neither. + #math.equation(block: false, alt: "3 , 5 , 7 , ⋯")[$3 , " " 5 , " " 7 , " " ⋯$] + #math.equation(block: false, alt: "3 , minus 6 , 12 , ⋯")[$3 , " " − 6 , " " 12 , " " ⋯$] + #math.equation(block: false, alt: "3 , 6 , 10 , ⋯")[$3 , " " 6 , " " 10 , " " ⋯$] + #math.equation(block: false, alt: "3 , 1 , the fraction 1 over 3 , ⋯")[$3 , " " 1 , " " display(frac(1, 3)) , ⋯$] #solutionbox[ + This sequence is arithmetic. Each term is obtained from the previous term by adding 2. + This sequence is geometric. Each term is obtained from the previous term by multiplying by #math.equation(block: false, alt: "minus 2")[$− 2$]. + This sequence is neither arithmetic or geometric. + This sequence is geometric. Each term is obtained from the previous term by multiplying by #math.equation(block: false, alt: "the fraction 1 over 3")[$display(frac(1, 3))$]. ] ] Identify each of the following sequences as arithmetic, geometric, or neither. + #math.equation(block: false, alt: "1 , 4 , 9 , 16 , ⋯")[$1 , " " 4 , " " 9 , " " 16 , " " ⋯$] \_\_\_\_\_ + #math.equation(block: false, alt: "2 , 6 , 18 , 54 , ⋯")[$2 , " " 6 , " " 18 , " " 54 , " " ⋯$] \_\_\_\_\_ + #math.equation(block: false, alt: "the fraction 1 over 3 , the fraction 1 over 2 , the fraction 2 over 3 , the fraction 5 over 6 , ⋯")[$display(frac(1, 3)) , " " display(frac(1, 2)) , " " display(frac(2, 3)) , " " display(frac(5, 6)) , " " ⋯$] \_\_\_\_\_ #solutionbox[ + Neither + Geometric + Arithmetic ] Identify each of the following sequences as arithmetic, geometric, or neither. + #math.equation(block: false, alt: "1 , 4 , 9 , 16 , ⋯")[$1 , " " 4 , " " 9 , " " 16 , " " ⋯$] + #math.equation(block: false, alt: "2 , 6 , 18 , 54 , ⋯")[$2 , " " 6 , " " 18 , " " 54 , " " ⋯$] + #math.equation(block: false, alt: "the fraction 1 over 3 , the fraction 1 over 2 , the fraction 2 over 3 , the fraction 5 over 6 , ⋯")[$display(frac(1, 3)) , " " display(frac(1, 2)) , " " display(frac(2, 3)) , " " display(frac(5, 6)) , " " ⋯$] === The General Term of a Geometric Sequence We can now find a non-recursive formula for the general term of a geometric sequence. If we denote the first term of the geometric sequence by #math.equation(block: false, alt: "a")[$a$], then the #math.equation(block: true, alt: "second term is, a r; third term is, a r times r equals a r squared; fourth term is, a r squared times r equals a r cubed; n to the power th term is, a r to the power n minus 1")[$"second term is" " " " " " " " " " " " " & a r \ "third term is" " " " " " " " " " " " " & a r ⋅ r = a r^(2) \ "fourth term is" " " " " " " " " " " " " & a r^(2) ⋅ r = a r^(3) \ n^("th") " " "term is" " " " " " " " " " " " " & a r^(n − 1)$] In general we have the following property. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Geometric Sequence] The #math.equation(block: false, alt: "n to the power th")[$n^("th")$] term of a geometric sequence is #math.equation(block: true, alt: "a sub n equals a r to the power n minus 1")[$a_(n) = a r^(n − 1)$] ] We can use the formula to find a particular term of a geometric sequence if we know the first term and the common ratio. #examplebox("Example 6")[][ Find the ninth term of the geometric sequence #math.equation(block: false, alt: "minus 24 , 12 , minus 6 , ⋯")[$− 24 , " " 12 , " " − 6 , " " ⋯$]. #solutionbox[ First, we find the common ratio by dividing any term by its predecessor. #math.equation(block: true, alt: "the fraction 12 over minus 24 equals the fraction minus 1 over 2")[$display(frac(12, − 24)) = display(frac(− 1, 2))$] Then we use the formula with #math.equation(block: false, alt: "a equals minus 24 , r equals the fraction minus 1 over 2")[$a = − 24 , " " r = display(frac(− 1, 2))$], and #math.equation(block: false, alt: "n equals 9")[$n = 9$]. #math.equation(block: true, alt: "a sub 9 equals minus 24 open parenthesis the fraction minus 1 over 2 close parenthesis to the power 8 equals the fraction minus 3 over 32")[$a_(9) = − 24 ( display(frac(− 1, 2)) )^(8) = display(frac(− 3, 32))$] ] ] Find the 12th term of the arithmetic sequence #math.equation(block: false, alt: "the fraction 3 over 32")[$#h(1em) display(frac(3, 32))$], #math.equation(block: false, alt: "the fraction 3 over 16")[$display(frac(3, 16))$], #math.equation(block: false, alt: "the fraction 3 over 8")[$display(frac(3, 8))$], #math.equation(block: false, alt: "the fraction 3 over 4 , and so on")[$display(frac(3, 4)) , …$] Answer: \_\_\_\_\_ #solutionbox[ 192 ] Find the 12th term of the arithmetic sequence #math.equation(block: false, alt: "the fraction 3 over 32")[$#h(1em) display(frac(3, 32))$], #math.equation(block: false, alt: "the fraction 3 over 16")[$display(frac(3, 16))$], #math.equation(block: false, alt: "the fraction 3 over 8")[$display(frac(3, 8))$], #math.equation(block: false, alt: "the fraction 3 over 4 , and so on")[$display(frac(3, 4)) , …$] #solutionbox[ 192 ] #examplebox("Example 7")[][ You deposit \$8000 in an account that pays 5% annual interest compounded annually. + Find a formula for the amount of money in the account after #math.equation(block: false, alt: "n")[$n$] years. + How much money will be in the account after 20 years? #solutionbox[ + Each year, the account balance is multiplied by 1.05. Thus, the annual balances #math.equation(block: false, alt: "c sub n")[$c_(n)$] form a geometric sequence whose first term is 8400 and whose common ratio is 1.05. Hence, #math.equation(block: true, alt: "c sub n equals 8400 open parenthesis 1.05 close parenthesis to the power n minus 1")[$c_(n) = 8400 ( 1.05 )^(n − 1)$] + We substitute #math.equation(block: false, alt: "n equals 20")[$n = 20$] into the formula for the general term found in part (a). #math.equation(block: true, alt: "c sub 20 equals 8400 open parenthesis 1.05 close parenthesis to the power 19 equals 21 , 226.38164")[$c_(20) = 8400 ( 1.05 )^(19) = 21 , 226.38164$] After 20 years, there will be \$21,226.38 in the account. ] ] Fill in the blanks. Use #math.equation(block: false, alt: "a")[$a$] for the first term of the sequence; use #math.equation(block: false, alt: "d")[$d$] for the common difference; use #math.equation(block: false, alt: "r")[$r$] for the common ratio. + The formula for the #math.equation(block: false, alt: "n")[$n$]th term of an arithmetic sequence is #linebreak() #math.equation(block: false, alt: "a sub n equals")[$a_(n) =$]\_\_\_\_\_ + The formula for the #math.equation(block: false, alt: "n")[$n$]th term of a geometric sequence is #linebreak() #math.equation(block: false, alt: "a sub n equals")[$a_(n) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "a sub n equals a plus open parenthesis n minus 1 close parenthesis d")[$a_(n) = a + ( n − 1 ) d$] + #math.equation(block: false, alt: "a sub n equals a r to the power n minus 1")[$a_(n) = a r^(n − 1)$] ] Fill in the blanks. + The formula for the #math.equation(block: false, alt: "n")[$n$]th term of an arithmetic sequence is \_\_\_\_\_\_\_\_\_\_ + The formula for the #math.equation(block: false, alt: "n")[$n$]th term of a geometric sequence is \_\_\_\_\_\_\_\_\_\_ Find a recursive definition for the sequence in the previous example. #math.equation(block: false, alt: "c sub 1 equals")[$c_(1) =$]\_\_\_\_\_, #math.equation(block: false, alt: "c sub n plus 1 equals")[$c_(n + 1) =$]\_\_\_\_\_ \[Note: Enter "cn" to get #math.equation(block: false, alt: "c sub n")[$c_(n)$].\] #solutionbox[ #math.equation(block: true, alt: "c sub 1 equals 8400")[$c_(1) = 8400$], #math.equation(block: true, alt: "c sub n plus 1 equals 1.05 c sub n")[$c_(n + 1) = 1.05 c_(n)$] ] Find a recursive definition for the sequence in the previous example, #math.equation(block: false, alt: "c sub n equals 8400 open parenthesis 1.05 close parenthesis to the power n minus 1")[$" " c_(n) = 8400 ( 1.05 )^(n − 1)$]. #solutionbox[ #math.equation(block: true, alt: "c sub 1 equals 8400")[$c_(1) = 8400$], #math.equation(block: true, alt: "c sub n plus 1 equals 1.05 c sub n")[$c_(n + 1) = 1.05 c_(n)$] ] #examplebox("Example 8")[][ Find a nonrecursive definition for each sequence. + #math.equation(block: false, alt: "a sub 1 equals 2 , a sub n equals a sub n minus 1 plus 3")[$a_(1) = 2 , " " " " a_(n) = a_(n − 1) + 3$] + #math.equation(block: false, alt: "b sub 1 equals 2 , b sub n plus 1 equals 3 b sub n")[$b_(1) = 2 , " " " " b_(n + 1) = 3 b_(n)$] #solutionbox[ + From the definition we see that the first few terms of the sequence are 2, 5, 8, and 11. Because each new term is found by adding 3 to the previous term, we see that this is an arithmetic sequence with a common difference of 3. This means that the sequence has the form #math.equation(block: true, alt: "a sub n equals 2 plus open parenthesis n minus 1 close parenthesis 3 or a sub n equals 3 n minus 1")[$a_(n) = 2 + ( n − 1 ) 3 " " " " " " "or" " " " " " " " " a_(n) = 3 n − 1$] + The first four terms of this sequence are 2, 6, 18, and 54. Because each new term is found by multiplying the previous term by 3, we see that this is a geometric sequence with a common ratio of 3. This means that the sequence has the form #math.equation(block: true, alt: "b sub n equals 3 times 2 to the power n minus 1")[$b_(n) = 3 ⋅ 2^(n − 1)$] ] ] Find a nonrecursive definition for each sequence. + #math.equation(block: false, alt: "c sub 1 equals 100")[$c_(1) = 100$], #math.equation(block: false, alt: "c sub k plus 1 equals 1.03 c sub k")[$#h(1em) c_(k + 1) = 1.03 c_(k)$] #linebreak() #math.equation(block: false, alt: "c sub k equals")[$c_(k) =$] \_\_\_\_\_ + #math.equation(block: false, alt: "d sub 1 equals 2")[$d_(1) = 2$], #math.equation(block: false, alt: "d sub k equals d sub k minus 1 minus 5")[$#h(1em) d_(k) = d_(k − 1) − 5$] #linebreak() #math.equation(block: false, alt: "d sub k equals")[$d_(k) =$] \_\_\_\_\_ S #solutionbox[ + #math.equation(block: false, alt: "c sub k equals 100 times 1.03 to the power k minus 1")[$c_(k) = 100 ⋅ 1.03^(k − 1)$] + #math.equation(block: false, alt: "d sub k equals 7 minus 5 k")[$d_(k) = 7 − 5 k$] ] Find a nonrecursive definition for each sequence. + #math.equation(block: false, alt: "c sub 1 equals 100")[$c_(1) = 100$], #math.equation(block: false, alt: "c sub k plus 1 equals 1.03 c sub k")[$#h(1em) c_(k + 1) = 1.03 c_(k)$] + #math.equation(block: false, alt: "d sub 1 equals 2")[$d_(1) = 2$], #math.equation(block: false, alt: "d sub k equals d sub k minus 1 minus 5")[$#h(1em) d_(k) = d_(k − 1) − 5$] #solutionbox[ + #math.equation(block: false, alt: "c sub k equals 100 times 1.03 to the power k minus 1")[$c_(k) = 100 ⋅ 1.03^(k − 1)$] + #math.equation(block: false, alt: "d sub k equals 7 minus 5 k")[$d_(k) = 7 − 5 k$] ] === Section Summary ==== Vocabulary - Arithmetic sequence - Common difference - General term - Geometric sequence - Common ratio - Recursive sequence ==== CONCEPTS + A sequence in which each term can be obtained from the previous term by adding a fixed amount is called an #strong[arithmetic sequence]. + The fixed amount between two successive terms of an arithmetic sequence is called the #strong[common difference]. + An arithmetic sequence defines a linear function of #math.equation(block: false, alt: "n")[$n$]. + The #math.equation(block: false, alt: "n to the power th")[$n^("th")$] term of an arithmetic sequence is #math.equation(block: false, alt: "a sub n equals a plus open parenthesis n minus 1 close parenthesis d")[$a_(n) = a + ( n − 1 ) d$]. + A sequence in which each term can be obtained from the previous term by multiplying by a fixed amount is called a #strong[geometric sequence]. + The ratio of two successive terms of a geometridc sequence is called the #strong[common ratio]. + A geometric sequence defines an exponential function of #math.equation(block: false, alt: "n")[$n$]. + The #math.equation(block: false, alt: "n to the power th")[$n^("th")$] term of a geometric sequence is #math.equation(block: false, alt: "a sub n equals a r to the power n minus 1")[$a_(n) = a r^(n − 1)$]. ==== STUDY QUESTIONS + What is an arithmetic sequence? What is the common difference? + State a formula for the general term of an arithemetic sequence. + An arithmetic sequence defines what kind of function of #math.equation(block: false, alt: "n")[$n$]? + What is a geometric sequence? What is the common ratio? + State a formula for the general term of a geometric sequence. + A geometric sequence defines what kind of function of #math.equation(block: false, alt: "n")[$n$]? ==== SKILLS Practice each skill in the Homework Problems listed. + Identify arithmetic and geometric sequences: \#1–10 + Evaluate the general term of an arithmetic or geometric sequence: \#13–24, \#33–36, + Write an expression for the general term: \#25–38 + Find the general term from a recursive definition: \#41–48 + Use a sequence as a model: \#49–56 === Homework 9.2 For Problems 1–12, identify the sequence as arithmetic, geometric, or neither. #math.equation(block: true, alt: "minus 2 , minus 6 , minus 18 , minus 54 , ⋯")[$− 2 , − 6 , − 18 , − 54 , ⋯$] Geometric #math.equation(block: true, alt: "minus 2 , minus 6 , minus 10 , minus 14 , ⋯")[$− 2 , − 6 , − 10 , − 14 , ⋯$] #math.equation(block: true, alt: "16 , 8 , 0 , minus 8 , ⋯")[$16 , 8 , 0 , − 8 , ⋯$] Arithmetic #math.equation(block: true, alt: "16 , 8 , 4 , 2 , ⋯")[$16 , 8 , 4 , 2 , ⋯$] #math.equation(block: true, alt: "minus 1 , 1 , minus 1 , 1 , ⋯")[$− 1 , 1 , − 1 , 1 , ⋯$] Geometric #math.equation(block: true, alt: "1 , 3 , 6 , 10 , ⋯")[$1 , 3 , 6 , 10 , ⋯$] #math.equation(block: true, alt: "1 , 4 , 9 , 16 , ⋯")[$1 , 4 , 9 , 16 , ⋯$] Neither #math.equation(block: true, alt: "5 , minus 5 , 5 , minus 5 , ⋯")[$5 , − 5 , 5 , − 5 , ⋯$] #math.equation(block: true, alt: "27 , 9 , 3 , 1 , ⋯")[$27 , 9 , 3 , 1 , ⋯$] Geometric #math.equation(block: true, alt: "the fraction minus 1 over 3 , the fraction 1 over 3 , 1 , the fraction 5 over 3 , ⋯")[$display(frac(− 1, 3)) , display(frac(1, 3)) , 1 , display(frac(5, 3)) , ⋯$] #math.equation(block: true, alt: "the fraction 2 over 3 , minus 1 , the fraction 3 over 2 , the fraction minus 9 over 4 , ⋯")[$display(frac(2, 3)) , − 1 , display(frac(3, 2)) , display(frac(− 9, 4)) , ⋯$] Geometric #math.equation(block: true, alt: "2 , minus 8 , 32 , minus 124 , ⋯")[$2 , − 8 , 32 , − 124 , ⋯$] For Problems 13–24, find the first four terms of the sequence. #math.equation(block: true, alt: "a equals 2 , d equals 4")[$a = 2 , d = 4$] #math.equation(block: true, alt: "2 , 6 , 10 , 14")[$2 , 6 , 10 , 14$] #math.equation(block: true, alt: "a equals 7 , d equals 3")[$a = 7 , d = 3$] #math.equation(block: true, alt: "a equals the fraction 1 over 2 , d equals the fraction 1 over 4")[$a = display(frac(1, 2)) , d = display(frac(1, 4))$] #math.equation(block: true, alt: "the fraction 1 over 2 , the fraction 3 over 4 , 1 , the fraction 5 over 4")[$display(frac(1, 2)) , display(frac(3, 4)) , 1 , display(frac(5, 4))$] #math.equation(block: true, alt: "a equals the fraction 2 over 3 , d equals the fraction 1 over 3")[$a = display(frac(2, 3)) , d = display(frac(1, 3))$] #math.equation(block: true, alt: "a equals 2.7 , d equals minus 0.8")[$a = 2.7 , d = − 0.8$] #math.equation(block: true, alt: "2.7 , 1.9 , 1.1 , 0.3")[$2.7 , 1.9 , 1.1 , 0.3$] #math.equation(block: true, alt: "a equals 5.9 , d equals minus 1.3")[$a = 5.9 , d = − 1.3$] #math.equation(block: true, alt: "a equals 5 , r equals minus 2")[$a = 5 , r = − 2$] #math.equation(block: true, alt: "5 , minus 10 , 20 , minus 40")[$5 , − 10 , 20 , − 40$] #math.equation(block: true, alt: "a equals minus 4 , r equals 3")[$a = − 4 , r = 3$] #math.equation(block: true, alt: "a equals 9 , r equals the fraction 2 over 3")[$a = 9 , r = display(frac(2, 3))$] #math.equation(block: true, alt: "9 , 6 , 4 , the fraction 8 over 3")[$9 , 6 , 4 , display(frac(8, 3))$] #math.equation(block: true, alt: "a equals 25 , r equals the fraction 4 over 5")[$a = 25 , r = display(frac(4, 5))$] #math.equation(block: true, alt: "a equals 60 , r equals 0.4")[$a = 60 , r = 0.4$] #math.equation(block: true, alt: "60 , 24 , 9.6 , 3.84")[$60 , 24 , 9.6 , 3.84$] #math.equation(block: true, alt: "a equals 100 , r equals 0.3")[$a = 100 , r = 0.3$] For Problems 25–32, find the next three terms, and an expression for the general term. #math.equation(block: true, alt: "3 , 7 , 11 , ⋯")[$3 , 7 , 11 , ⋯$] #math.equation(block: true, alt: "15 , 19 , 23 , ⋯ , 3 plus open parenthesis n minus 1 close parenthesis 4")[$15 , 19 , 23 , ⋯ , 3 + ( n − 1 ) 4$] #math.equation(block: true, alt: "minus 10 , minus 20 , minus 30 , ⋯")[$− 10 , − 20 , − 30 , ⋯$] #math.equation(block: true, alt: "minus 1 , minus 5 , minus 9 , ⋯")[$− 1 , − 5 , − 9 , ⋯$] #math.equation(block: true, alt: "minus 13 , minus 17 , minus 21 , ⋯ , minus 1 minus 4 open parenthesis n minus 1 close parenthesis")[$− 13 , − 17 , − 21 , ⋯ , − 1 − 4 ( n − 1 )$] #math.equation(block: true, alt: "minus 6 , minus 1 , 4 , ⋯")[$− 6 , − 1 , 4 , ⋯$] #math.equation(block: true, alt: "the fraction 2 over 3 , the fraction 4 over 3 , the fraction 8 over 3 , ⋯")[$display(frac(2, 3)) , display(frac(4, 3)) , display(frac(8, 3)) , ⋯$] #math.equation(block: true, alt: "the fraction 16 over 3 , the fraction 32 over 3 , the fraction 64 over 3 , ⋯ , the fraction 2 over 3 open parenthesis 2 close parenthesis to the power n minus 1")[$display(frac(16, 3)) , display(frac(32, 3)) , display(frac(64, 3)) , ⋯ , display(frac(2, 3)) ( 2 )^(n − 1)$] #math.equation(block: true, alt: "6 , 3 , the fraction 3 over 2 , ⋯")[$6 , 3 , display(frac(3, 2)) , ⋯$] #math.equation(block: true, alt: "4 , minus 2 , 1 , ⋯")[$4 , − 2 , 1 , ⋯$] #math.equation(block: true, alt: "the fraction minus 1 over 2 , the fraction 1 over 4 , the fraction minus 1 over 8 , ⋯ , 4 open parenthesis the fraction minus 1 over 2 close parenthesis to the power n minus 1")[$display(frac(− 1, 2)) , display(frac(1, 4)) , display(frac(− 1, 8)) , ⋯ , 4 ( display(frac(− 1, 2)) )^(n − 1)$] #math.equation(block: true, alt: "the fraction 1 over 2 , the fraction minus 3 over 2 , the fraction 9 over 2 , ⋯")[$display(frac(1, 2)) , display(frac(− 3, 2)) , display(frac(9, 2)) , ⋯$] Find the twelfth term in the arithmetic sequence #math.equation(block: false, alt: "2 , the fraction 5 over 2 , 3 , ⋯")[$2 , display(frac(5, 2)) , 3 , ⋯$] #math.equation(block: true, alt: "7.5")[$7.5$] Find the tenth term in the arithmetic sequence #math.equation(block: false, alt: "the fraction 3 over 2 , 2 , the fraction 13 over 4 , ⋯")[$display(frac(3, 2)) , 2 , display(frac(13, 4)) , ⋯$] Find the eighth term in the geometric sequence #math.equation(block: false, alt: "minus 3 , the fraction 3 over 2 , the fraction minus 3 over 4 , ⋯")[$− 3 , display(frac(3, 2)) , display(frac(− 3, 4)) , ⋯$] #math.equation(block: true, alt: "the fraction 3 over 128")[$display(frac(3, 128))$] Find the sixth term in the geometric sequence #math.equation(block: false, alt: "minus 5 , 1 , the fraction minus 1 over 5 , ⋯")[$− 5 , 1 , display(frac(− 1, 5)) , ⋯$] Find the first term of a geometric sequence with fifth term #math.equation(block: false, alt: "48")[$48$] and common ratio #math.equation(block: false, alt: "2")[$2$]. #math.equation(block: true, alt: "3")[$3$] Find the first term of a geometric sequence with fifth term #math.equation(block: false, alt: "1")[$1$] and common ratio #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]. How many terms are in the sequence #math.equation(block: false, alt: "the fraction 1 over 8 , the fraction 1 over 4 , the fraction 1 over 2 , ⋯ , 512")[$display(frac(1, 8)) , display(frac(1, 4)) , display(frac(1, 2)) , ⋯ , 512$]? #math.equation(block: true, alt: "13")[$13$] How many terms are in the sequence #math.equation(block: false, alt: "the fraction 27 over 64 , the fraction 9 over 16 , the fraction 3 over 4 , ⋯ , the fraction 64 over 27")[$display(frac(27, 64)) , display(frac(9, 16)) , display(frac(3, 4)) , ⋯ , display(frac(64, 27))$]? For Problems 41–48, find a nonrecursive definition for the sequence. #math.equation(block: true, alt: "s sub 1 equals 3 , s sub n equals s sub n minus 1 plus 2")[$s_(1) = 3 , " " " " s_(n) = s_(n − 1) + 2$] #math.equation(block: true, alt: "s sub n equals 3 plus 2 open parenthesis n minus 1 close parenthesis")[$s_(n) = 3 + 2 ( n − 1 )$] #math.equation(block: true, alt: "c sub 1 equals 6 , c sub n equals c sub n minus 1 minus 4")[$c_(1) = 6 , " " " " c_(n) = c_(n − 1) − 4$] #math.equation(block: true, alt: "x sub 1 equals 0 , x sub n plus 1 equals x sub n minus 3")[$x_(1) = 0 , " " " " x_(n + 1) = x_(n) − 3$] #math.equation(block: true, alt: "x sub n equals minus 3 open parenthesis n minus 1 close parenthesis")[$x_(n) = − 3 ( n − 1 )$] #math.equation(block: true, alt: "y sub 1 equals minus 1 , y sub n plus 1 equals y sub n plus 5")[$y_(1) = − 1 , " " " " y_(n + 1) = y_(n) + 5$] #math.equation(block: true, alt: "d sub 1 equals 24 , d sub n plus 1 equals the fraction minus 1 over 2 d sub n")[$d_(1) = 24 , " " " " d_(n + 1) = display(frac(− 1, 2)) d_(n)$] #math.equation(block: true, alt: "d sub n equals 24 open parenthesis the fraction minus 1 over 2 close parenthesis to the power n minus 1")[$d_(n) = 24 ( display(frac(− 1, 2)) )^(n − 1)$] #math.equation(block: true, alt: "r sub 1 equals 27 , r sub n plus 1 equals the fraction 2 over 3 r sub n")[$r_(1) = 27 , " " " " r_(n + 1) = display(frac(2, 3)) r_(n)$] #math.equation(block: true, alt: "w sub 1 equals 1 ; w sub n equals 2 w sub n minus 1")[$w_(1) = 1 ; " " " " w_(n) = 2 w_(n − 1)$] #math.equation(block: true, alt: "w sub n equals 2 to the power n minus 1")[$w_(n) = 2^(n − 1)$] #math.equation(block: true, alt: "q sub 1 equals 7 ; q sub n equals 3 q sub n minus 1")[$q_(1) = 7 ; " " " " q_(n) = 3 q_(n − 1)$] For Problems 49–56, + Write a formula for the general term of the sequence. + Answer the question in the problem. An outdoor theater has 30 seats in the first row, 32 seats in the second row, 34 seats in the third row, and so on, with two more seats in any one row than in the previous. Let #math.equation(block: false, alt: "s sub n")[$s_(n)$] be the number of seats in the #math.equation(block: false, alt: "n to the power th")[$n^("th")$] row of the theater. How many seats are there in the fiftieth row? + #math.equation(block: false, alt: "s sub n equals 30 plus 2 open parenthesis n minus 1 close parenthesis")[$s_(n) = 30 + 2 ( n − 1 )$] + 128 Gwynn is training for a triathlon. In her first week of training, she bicycled a total of 20 miles, then increased to 24 miles in her second week, 28 miles in her third week, and so on, increasing her mileage by 4 miles per week. Let #math.equation(block: false, alt: "m sub n")[$m_(n)$] be Gwynn's mileage in her #math.equation(block: false, alt: "n to the power th")[$n^("th")$] week of training. How many miles will she bicycle in her eighteenth week of training? The cost of drilling a well increases the deeper you drill. Mel's Wells charges \$50 for the first five feet, \$55 for the second five feet, \$60 for the third five feet, and so on. Let #math.equation(block: false, alt: "d sub n")[$d_(n)$] be the cost of drilling the #math.equation(block: false, alt: "n to the power th")[$n^("th")$] five feet. How much does Mel charge to drill from a depth of 65 feet to a depth of 70 feet? + #math.equation(block: false, alt: "d sub n equals 50 plus 5 open parenthesis n minus 1 close parenthesis")[$d_(n) = 50 + 5 ( n − 1 )$] + \$115 The cost of having the windows washed in a high-rise office building depends on the height of the window. We Do Windows charges \$0.50 per square yard of glass below 10 feet high, \$0.60 per square yard for windows between 10 and 20 feet above the ground, \$0.70 per square yard for windows between 20 and 30 feet above the ground, and so on. Let #math.equation(block: false, alt: "w sub n")[$w_(n)$] be the cost of washing windows on the #math.equation(block: false, alt: "n to the power th")[$n^("th")$] floor if each floor is 10 feet high. How much does We Do Windows charge to wash 100 square yards of windows on the twenty-third floor? Valerie's grandmother deposited \$500 into a college fund for Valerie on the day she was born. The fund earns 5% interest compounded annually. Let #math.equation(block: false, alt: "V sub n")[$V_(n)$] be the value of the deposit, including interest, #math.equation(block: false, alt: "n")[$n$] years later. How much will the deposit be worth on Valerie's eighteenth birthday? + #math.equation(block: false, alt: "V sub n equals 500 open parenthesis 1.05 close parenthesis to the power n")[$V_(n) = 500 ( 1.05 )^(n)$] + \$1203.31 When he was 25 years old, Bruce won \$2000 on the lottery and deposited the money into a retirement fund. The fund earns 6% interest compounded annually. Let #math.equation(block: false, alt: "R sub n")[$R_(n)$] be the value of the deposit, including interest, #math.equation(block: false, alt: "n")[$n$] years later. How much will the deposit be worth when Bruce turns 65 years old? One hundred kilograms of a toxic chemical was dumped illegally into a clean reservoir. A filter can remove 20% of the chemical still present each week (so that 80% of the previous amount remains.) Let #math.equation(block: false, alt: "c sub n")[$c_(n)$] be the amount of the chemical remaining after #math.equation(block: false, alt: "n")[$n$] weeks. How much of the chemical will remain in the water after 20 weeks? + #math.equation(block: false, alt: "c sub n equals 100 open parenthesis 0.8 close parenthesis to the power n")[$c_(n) = 100 ( 0.8 )^(n)$] + 1.15 kg A heart patient is given 40 milliliters of a medication by injection. Each hour, 15% of the medicine still present is eliminated from the body (so that 85% of the previous amount remains.) Let #math.equation(block: false, alt: "m sub n")[$m_(n)$] be the amount of medicine remaining in the patient's body after #math.equation(block: false, alt: "n")[$n$] hours. How much of the medication is left in the patient's body after 10 hours?