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IB Mathematics: Applications and Interpretation HL · 12.1: Sequences & Series · 12.1.10
Sequences and series: revision Problem set A One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
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The sum of the first n terms of an arithmetic sequence is \(S_n=4n^2-n\).
(a) Write down the values of \(S_1\) and \(S_2\).
- \(S_1=4(1)^2-1(1)=3\)
- \(S_2=4(2)^2-1(2)=14\)
(b) Write down the values of \(u_1\) and \(u_2\).
- \(u_1=S_1=3\)
- \(u_2=S_2-S_1=14-3=11\)
(c) Find d, the common difference.
- \(d=u_2-u_1=11-3=8\)
(d) Find \(u_{15}\).
- \(u_n=3+(n-1)(8)=8n-5\)
- \(u_{15}=3+14(8)=115\)
(e) Find the greatest value of n for which \(u_n<150\).
- \(8n-5<150\)
- From the calculator: \(n<19.4\)
- The greatest whole number is \(n=19\)
(f) Find the value of n for which \(S_n=1278\).
- \(4n^2-n=1278\)
- From the calculator: \(n=18\) or \(n=−17.8\)
- n is a positive whole number: \(n=18\)
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The term \(u_5\) of an arithmetic sequence is 40 and the common difference is \(-3\). Find \(u_1\).
- \(u_5=u_1+4d\Rightarrow 40=u_1+4(-3)\)
- \(u_1=40+12=52\)
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An arithmetic sequence has \(u_{6}=17\) and \(u_{14}=49\).
(a) Find the common difference.
- \(u_1+5d=17\) and \(u_1+13d=49\)
- Subtract: \(8d=32\Rightarrow d=4\)
(b) Find the first term.
- \(u_1=17-5(4)=-3\)
(c) Find \(u_{60}\).
- \(u_{60}=-3+59(4)=233\)
(d) Find \(S_{100}\).
- \(S_{100}=\frac{100}{2}\left(2(-3)+99(4)\right)\)
- \(S_{100}=19500\)
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The term \(u_4\) of an arithmetic sequence is 20 and the common difference is \(-4\). The first and third terms of this sequence are the first two terms of a geometric sequence. Find the sixth term of the geometric sequence.
- \(u_1=20-3(-4)=32\), \(u_3=32+2(-4)=24\)
- \(r=\frac{24}{32}=\frac{3}{4}\)
- \(u_6=32\left(\frac{3}{4}\right)^5=7.59\)
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The term \(u_2\) of an arithmetic sequence is 22 and the common difference is \(-6\). The first and fourth terms of this sequence are the first two terms of an infinite geometric series. Find the sum to infinity of the geometric series.
- \(u_1=22-1(-6)=28\), \(u_4=28+3(-6)=10\)
- \(r=\frac{10}{28}=\frac{5}{14}\), and \(|r|<1\)
- \(S_\infty=\frac{28}{1-\frac{5}{14}}=\frac{392}{9}=43.6\)
In context
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A theatre has 20 seats in the first row. Each row after that has 3 more seats than the row in front.
(a) Find the number of seats in the 25th row.
- \(u_1=20,\ d=3\)
- \(u_{25}=20+24(3)=92\)
(b) Find the total number of seats in the first 25 rows.
- \(S_{25}=\frac{25}{2}\left(2(20)+24(3)\right)=1400\)
(c) The theatre needs at least 2000 seats. Find the least number of rows it must have.
- \(S_n=\frac n2\left(2(20)+(n-1)(3)\right)\ge2000\)
- From the calculator: \(S_n=2000\) at \(n=30.9\)
- The least whole number is \(n=31\)
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A factory makes 800 units in its first month. Each month it makes 5% more units than the month before. Give your answers to the nearest whole number.
(a) Find the number of units made in month 12.
- Geometric: \(u_1=800,\ r=1.05\)
- \(u_{12}=800(1.05)^{11}=1368\)
(b) Find the total number of units made in the first 12 months.
- \(S_{12}=\frac{800\left(1.05^{12}-1\right)}{1.05-1}=12734\)
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A patient takes 60 mg of a drug on day 1. Each later day the dose is 80% of the dose the day before, and this continues for ever. Find the total amount of the drug taken.
- \(u_1=60,\ r=0.8\), and \(|r|<1\)
- \(S_\infty=\frac{60}{1-0.8}=300\)
Stretch
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An arithmetic sequence has \(S_{10}=215\) and \(S_{20}=830\). Find the first term and the common difference.
- \(\frac{10}{2}(2u_1+9d)=215\), so \(2u_1+9d=43\)
- \(\frac{20}{2}(2u_1+19d)=830\), so \(2u_1+19d=83\)
- Subtract: \(10d=40\), so \(d=4\)
- \(2u_1=43-9(4)\), so \(u_1=3.5\)
jmaths.xyz · AI HL · 12.1.10 Sequences and series: revision