IB AI · Unit 2 · 2A Arithmetic Sequences
Some sums are arithmetic lists with a rule, such as selected multiples or pieces cut from one length.
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Multiples form an arithmetic sequence
List the first multiple, the last multiple, and the gap between multiples. That gap is d. Then
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
Take out a second list
To skip the multiples that also belong to another list, add the full list and subtract the overlap. The overlap is a new arithmetic sequence.
Pieces of one total
The lengths are the terms, and the whole length is Sₙ. If the longest piece is k times the shortest piece, then uₙ = k u₁.
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
Add every term in one arithmetic list.
These numbers are an arithmetic sequence with first term 6, common difference 3, and last term 36.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{36 - 6}{3} + 1\)
\(n = \dfrac{30}{3} + 1\)
\(n = 10 + 1\)
\(n = 11\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{11} = \dfrac{11}{2}\left(6 + 36\right)\)
\(6 + 36 = 42\)
\(S_{11} = \dfrac{11}{2} \times 42\)
\(S_{11} = 11 \times 21\)
\(S_{11} = 231\)
Answer: the sum = \(231\)
These numbers are an arithmetic sequence with first term 10, common difference 5, and last term 60.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{60 - 10}{5} + 1\)
\(n = \dfrac{50}{5} + 1\)
\(n = 10 + 1\)
\(n = 11\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{11} = \dfrac{11}{2}\left(10 + 60\right)\)
\(10 + 60 = 70\)
\(S_{11} = \dfrac{11}{2} \times 70\)
\(S_{11} = 11 \times 35\)
\(S_{11} = 385\)
Answer: the sum = \(385\)
These numbers are an arithmetic sequence with first term 8, common difference 4, and last term 48.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{48 - 8}{4} + 1\)
\(n = \dfrac{40}{4} + 1\)
\(n = 10 + 1\)
\(n = 11\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{11} = \dfrac{11}{2}\left(8 + 48\right)\)
\(8 + 48 = 56\)
\(S_{11} = \dfrac{11}{2} \times 56\)
\(S_{11} = 11 \times 28\)
\(S_{11} = 308\)
Answer: the sum = \(308\)
These numbers are an arithmetic sequence with first term 6, common difference 6, and last term 66.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{66 - 6}{6} + 1\)
\(n = \dfrac{60}{6} + 1\)
\(n = 10 + 1\)
\(n = 11\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{11} = \dfrac{11}{2}\left(6 + 66\right)\)
\(6 + 66 = 72\)
\(S_{11} = \dfrac{11}{2} \times 72\)
\(S_{11} = 11 \times 36\)
\(S_{11} = 396\)
Answer: the sum = \(396\)
These numbers are an arithmetic sequence with first term 18, common difference 9, and last term 81.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{81 - 18}{9} + 1\)
\(n = \dfrac{63}{9} + 1\)
\(n = 7 + 1\)
\(n = 8\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{8} = \dfrac{8}{2}\left(18 + 81\right)\)
\(18 + 81 = 99\)
\(S_{8} = \dfrac{8}{2} \times 99\)
\(S_{8} = 4 \times 99\)
\(S_{8} = 396\)
Answer: the sum = \(396\)
These numbers are an arithmetic sequence with first term 1, common difference 1, and last term 50.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{50 - 1}{1} + 1\)
\(n = \dfrac{49}{1} + 1\)
\(n = 49 + 1\)
\(n = 50\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{50} = \dfrac{50}{2}\left(1 + 50\right)\)
\(1 + 50 = 51\)
\(S_{50} = \dfrac{50}{2} \times 51\)
\(S_{50} = 25 \times 51\)
\(S_{50} = 1275\)
Answer: the sum = \(1275\)
Add one list of multiples, then subtract the overlap.
Add the multiples of 5 from 0 through 60.
These numbers are an arithmetic sequence with first term 0, common difference 5, and last term 60.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{60 - 0}{5} + 1\)
\(n = \dfrac{60}{5} + 1\)
\(n = 12 + 1\)
\(n = 13\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{13} = \dfrac{13}{2}\left(0 + 60\right)\)
\(0 + 60 = 60\)
\(S_{13} = \dfrac{13}{2} \times 60\)
\(S_{13} = 13 \times 30\)
\(S_{13} = 390\)
Subtract the multiples of 15, because those are multiples of both 5 and 3.
These numbers are an arithmetic sequence with first term 0, common difference 15, and last term 60.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{60 - 0}{15} + 1\)
\(n = \dfrac{60}{15} + 1\)
\(n = 4 + 1\)
\(n = 5\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{5} = \dfrac{5}{2}\left(0 + 60\right)\)
\(0 + 60 = 60\)
\(S_{5} = \dfrac{5}{2} \times 60\)
\(S_{5} = 5 \times 30\)
\(S_{5} = 150\)
Subtract the two sums.
\(390 - 150 = 240\)
Answer: the sum = \(240\)
Add the multiples of 5 from 0 through 45.
These numbers are an arithmetic sequence with first term 0, common difference 5, and last term 45.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{45 - 0}{5} + 1\)
\(n = \dfrac{45}{5} + 1\)
\(n = 9 + 1\)
\(n = 10\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{10} = \dfrac{10}{2}\left(0 + 45\right)\)
\(0 + 45 = 45\)
\(S_{10} = \dfrac{10}{2} \times 45\)
\(S_{10} = 5 \times 45\)
\(S_{10} = 225\)
Subtract the multiples of 15, because those are multiples of both 5 and 3.
These numbers are an arithmetic sequence with first term 0, common difference 15, and last term 45.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{45 - 0}{15} + 1\)
\(n = \dfrac{45}{15} + 1\)
\(n = 3 + 1\)
\(n = 4\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{4} = \dfrac{4}{2}\left(0 + 45\right)\)
\(0 + 45 = 45\)
\(S_{4} = \dfrac{4}{2} \times 45\)
\(S_{4} = 2 \times 45\)
\(S_{4} = 90\)
Subtract the two sums.
\(225 - 90 = 135\)
Answer: the sum = \(135\)
Add the multiples of 4 from 0 through 48.
These numbers are an arithmetic sequence with first term 0, common difference 4, and last term 48.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{48 - 0}{4} + 1\)
\(n = \dfrac{48}{4} + 1\)
\(n = 12 + 1\)
\(n = 13\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{13} = \dfrac{13}{2}\left(0 + 48\right)\)
\(0 + 48 = 48\)
\(S_{13} = \dfrac{13}{2} \times 48\)
\(S_{13} = 13 \times 24\)
\(S_{13} = 312\)
Subtract the multiples of 12, because those are multiples of both 4 and 3.
These numbers are an arithmetic sequence with first term 0, common difference 12, and last term 48.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{48 - 0}{12} + 1\)
\(n = \dfrac{48}{12} + 1\)
\(n = 4 + 1\)
\(n = 5\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{5} = \dfrac{5}{2}\left(0 + 48\right)\)
\(0 + 48 = 48\)
\(S_{5} = \dfrac{5}{2} \times 48\)
\(S_{5} = 5 \times 24\)
\(S_{5} = 120\)
Subtract the two sums.
\(312 - 120 = 192\)
Answer: the sum = \(192\)
Add the multiples of 6 from 0 through 72.
These numbers are an arithmetic sequence with first term 0, common difference 6, and last term 72.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{72 - 0}{6} + 1\)
\(n = \dfrac{72}{6} + 1\)
\(n = 12 + 1\)
\(n = 13\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{13} = \dfrac{13}{2}\left(0 + 72\right)\)
\(0 + 72 = 72\)
\(S_{13} = \dfrac{13}{2} \times 72\)
\(S_{13} = 13 \times 36\)
\(S_{13} = 468\)
Subtract the multiples of 12, because those are multiples of both 6 and 4.
These numbers are an arithmetic sequence with first term 0, common difference 12, and last term 72.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{72 - 0}{12} + 1\)
\(n = \dfrac{72}{12} + 1\)
\(n = 6 + 1\)
\(n = 7\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{7} = \dfrac{7}{2}\left(0 + 72\right)\)
\(0 + 72 = 72\)
\(S_{7} = \dfrac{7}{2} \times 72\)
\(S_{7} = 7 \times 36\)
\(S_{7} = 252\)
Subtract the two sums.
\(468 - 252 = 216\)
Answer: the sum = \(216\)
Add the multiples of 5 from 0 through 80.
These numbers are an arithmetic sequence with first term 0, common difference 5, and last term 80.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{80 - 0}{5} + 1\)
\(n = \dfrac{80}{5} + 1\)
\(n = 16 + 1\)
\(n = 17\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{17} = \dfrac{17}{2}\left(0 + 80\right)\)
\(0 + 80 = 80\)
\(S_{17} = \dfrac{17}{2} \times 80\)
\(S_{17} = 17 \times 40\)
\(S_{17} = 680\)
Subtract the multiples of 10, because those are multiples of both 5 and 2.
These numbers are an arithmetic sequence with first term 0, common difference 10, and last term 80.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{80 - 0}{10} + 1\)
\(n = \dfrac{80}{10} + 1\)
\(n = 8 + 1\)
\(n = 9\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{9} = \dfrac{9}{2}\left(0 + 80\right)\)
\(0 + 80 = 80\)
\(S_{9} = \dfrac{9}{2} \times 80\)
\(S_{9} = 9 \times 40\)
\(S_{9} = 360\)
Subtract the two sums.
\(680 - 360 = 320\)
Answer: the sum = \(320\)
Add the multiples of 7 from 0 through 84.
These numbers are an arithmetic sequence with first term 0, common difference 7, and last term 84.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{84 - 0}{7} + 1\)
\(n = \dfrac{84}{7} + 1\)
\(n = 12 + 1\)
\(n = 13\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{13} = \dfrac{13}{2}\left(0 + 84\right)\)
\(0 + 84 = 84\)
\(S_{13} = \dfrac{13}{2} \times 84\)
\(S_{13} = 13 \times 42\)
\(S_{13} = 546\)
Subtract the multiples of 21, because those are multiples of both 7 and 3.
These numbers are an arithmetic sequence with first term 0, common difference 21, and last term 84.
\(n = \dfrac{u_n - u_1}{d} + 1\)
\(n = \dfrac{84 - 0}{21} + 1\)
\(n = \dfrac{84}{21} + 1\)
\(n = 4 + 1\)
\(n = 5\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{5} = \dfrac{5}{2}\left(0 + 84\right)\)
\(0 + 84 = 84\)
\(S_{5} = \dfrac{5}{2} \times 84\)
\(S_{5} = 5 \times 42\)
\(S_{5} = 210\)
Subtract the two sums.
\(546 - 210 = 336\)
Answer: the sum = \(336\)
The pieces form an arithmetic sequence, and their lengths add to the original length.
Let the shortest piece be \(u_1\) and the longest piece be \(u_n\).
\(u_{6} = 2u_1\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(90 = \dfrac{6}{2}\left(u_1 + 2u_1\right)\)
\(90 = \dfrac{6}{2}\left(3\,u_1\right)\)
\(90 = 9\,u_1\)
\(u_1 = \dfrac{90}{9}\)
\(u_1 = 10\)
Answer: the shortest piece = \(10\) cm
Let the shortest piece be \(u_1\) and the longest piece be \(u_n\).
\(u_{8} = 3u_1\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(160 = \dfrac{8}{2}\left(u_1 + 3u_1\right)\)
\(160 = \dfrac{8}{2}\left(4\,u_1\right)\)
\(160 = 16\,u_1\)
\(u_1 = \dfrac{160}{16}\)
\(u_1 = 10\)
Answer: the shortest piece = \(10\) cm
Let the shortest piece be \(u_1\) and the longest piece be \(u_n\).
\(u_{10} = 4u_1\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(150 = \dfrac{10}{2}\left(u_1 + 4u_1\right)\)
\(150 = \dfrac{10}{2}\left(5\,u_1\right)\)
\(150 = 25\,u_1\)
\(u_1 = \dfrac{150}{25}\)
\(u_1 = 6\)
Answer: the shortest piece = \(6\) cm
Let the shortest piece be \(u_1\) and the longest piece be \(u_n\).
\(u_{5} = 2u_1\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(3 = \dfrac{5}{2}\left(u_1 + 2u_1\right)\)
\(3 = \dfrac{5}{2}\left(3\,u_1\right)\)
\(3 = \dfrac{15}{2}\,u_1\)
\(u_1 = 3 \div \dfrac{15}{2}\)
\(u_1 = 3 \times \dfrac{2}{15}\)
\(u_1 = \dfrac{6}{15}\)
\(u_1 = \dfrac{2}{5}\)
Answer: the shortest piece = \(\dfrac{2}{5}\) m
Let the shortest piece be \(u_1\) and the longest piece be \(u_n\).
\(u_{12} = 3u_1\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(96 = \dfrac{12}{2}\left(u_1 + 3u_1\right)\)
\(96 = \dfrac{12}{2}\left(4\,u_1\right)\)
\(96 = 24\,u_1\)
\(u_1 = \dfrac{96}{24}\)
\(u_1 = 4\)
Answer: the shortest piece = \(4\) cm
Let the shortest piece be \(u_1\) and the longest piece be \(u_n\).
\(u_{9} = 5u_1\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(180 = \dfrac{9}{2}\left(u_1 + 5u_1\right)\)
\(180 = \dfrac{9}{2}\left(6\,u_1\right)\)
\(180 = 27\,u_1\)
\(u_1 = \dfrac{180}{27}\)
\(u_1 = \dfrac{20}{3}\)
The longest piece is also the last term, so use the nth-term formula to find d.
\(\dfrac{100}{3} = \dfrac{20}{3} + (9-1)d\)
\(\dfrac{100}{3} - \dfrac{20}{3} = 8d\)
\(\dfrac{80}{3} = 8d\)
\(d = \dfrac{80}{24}\)
\(d = \dfrac{10}{3}\)
Answer: the shortest piece = \(\dfrac{20}{3}\) cm and the common difference = \(\dfrac{10}{3}\) cm
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