IB AI · Unit 2 · 2A Arithmetic Sequences
Substitute the values you know into the nth-term formula, then solve for the one value the question asks for.
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SL 1.2
\(u_n = u_1 + (n-1)d\)
\(S_n = \dfrac{n}{2}\bigl(2u_1 + (n-1)d\bigr)\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
The nth term
\(u_n = u_1 + (n-1)d\)
\(u_n\) = last term
\(u_1\) = first term
\(n\) = number of terms
\(d\) = common difference
Example. \(u_1 = 5\), \(d = 4\). Find \(u_{12}\).
\(u_{12}\) = the term to find
\(u_1 = 5\)
\(n = 12\)
\(d = 4\)
\(u_{12} = 5 + (12-1) \times 4\)
\(u_{12} = 5 + 11 \times 4\)
\(u_{12} = 5 + 44\)
\(u_{12} = 49\)
The first term and the common difference are given.
\(u_n = u_1 + (n-1)d\)
\(u_{10} = \text{the term to find}\)
\(u_1 = 7\)
\(n = 10\)
\(d = 3\)
\(u_{10} = 7 + (10-1) \times 3\)
\(u_{10} = 7 + 9 \times 3\)
\(u_{10} = 7 + 27\)
\(u_{10} = 34\)
Answer: the 10th term = \(34\)
\(u_n = u_1 + (n-1)d\)
\(u_{9} = \text{the term to find}\)
\(u_1 = 4\)
\(n = 9\)
\(d = 6\)
\(u_{9} = 4 + (9-1) \times 6\)
\(u_{9} = 4 + 8 \times 6\)
\(u_{9} = 4 + 48\)
\(u_{9} = 52\)
Answer: the 9th term = \(52\)
\(u_n = u_1 + (n-1)d\)
\(u_{6} = \text{the term to find}\)
\(u_1 = 11\)
\(n = 6\)
\(d = 4\)
\(u_{6} = 11 + (6-1) \times 4\)
\(u_{6} = 11 + 5 \times 4\)
\(u_{6} = 11 + 20\)
\(u_{6} = 31\)
Answer: the 6th term = \(31\)
\(u_n = u_1 + (n-1)d\)
\(u_{8} = \text{the term to find}\)
\(u_1 = 2\)
\(n = 8\)
\(d = 9\)
\(u_{8} = 2 + (8-1) \times 9\)
\(u_{8} = 2 + 7 \times 9\)
\(u_{8} = 2 + 63\)
\(u_{8} = 65\)
Answer: the 8th term = \(65\)
\(u_n = u_1 + (n-1)d\)
\(u_{20} = \text{the term to find}\)
\(u_1 = 15\)
\(n = 20\)
\(d = 2\)
\(u_{20} = 15 + (20-1) \times 2\)
\(u_{20} = 15 + 19 \times 2\)
\(u_{20} = 15 + 38\)
\(u_{20} = 53\)
Answer: the 20th term = \(53\)
\(u_n = u_1 + (n-1)d\)
\(u_{7} = \text{the term to find}\)
\(u_1 = 8\)
\(n = 7\)
\(d = 5\)
\(u_{7} = 8 + (7-1) \times 5\)
\(u_{7} = 8 + 6 \times 5\)
\(u_{7} = 8 + 30\)
\(u_{7} = 38\)
Answer: the 7th term = \(38\)
The first term, the term number, and that term are given.
\(u_n = u_1 + (n-1)d\)
\(u_{8} = 33\)
\(u_1 = 5\)
\(n = 8\)
\(d = \text{the common difference}\)
\(33 = 5 + (8-1)d\)
\(33 = 5 + 7d\)
\(33 - 5 = 7d\)
\(28 = 7d\)
\(d = \dfrac{28}{7}\)
\(d = 4\)
Answer: the common difference = \(4\)
\(u_n = u_1 + (n-1)d\)
\(u_{10} = 38\)
\(u_1 = 2\)
\(n = 10\)
\(d = \text{the common difference}\)
\(38 = 2 + (10-1)d\)
\(38 = 2 + 9d\)
\(38 - 2 = 9d\)
\(36 = 9d\)
\(d = \dfrac{36}{9}\)
\(d = 4\)
Answer: the common difference = \(4\)
\(u_n = u_1 + (n-1)d\)
\(u_{6} = 32\)
\(u_1 = 7\)
\(n = 6\)
\(d = \text{the common difference}\)
\(32 = 7 + (6-1)d\)
\(32 = 7 + 5d\)
\(32 - 7 = 5d\)
\(25 = 5d\)
\(d = \dfrac{25}{5}\)
\(d = 5\)
Answer: the common difference = \(5\)
\(u_n = u_1 + (n-1)d\)
\(u_{12} = 70\)
\(u_1 = 15\)
\(n = 12\)
\(d = \text{the common difference}\)
\(70 = 15 + (12-1)d\)
\(70 = 15 + 11d\)
\(70 - 15 = 11d\)
\(55 = 11d\)
\(d = \dfrac{55}{11}\)
\(d = 5\)
Answer: the common difference = \(5\)
\(u_n = u_1 + (n-1)d\)
\(u_{9} = 52\)
\(u_1 = 4\)
\(n = 9\)
\(d = \text{the common difference}\)
\(52 = 4 + (9-1)d\)
\(52 = 4 + 8d\)
\(52 - 4 = 8d\)
\(48 = 8d\)
\(d = \dfrac{48}{8}\)
\(d = 6\)
Answer: the common difference = \(6\)
\(u_n = u_1 + (n-1)d\)
\(u_{7} = 2\)
\(u_1 = 20\)
\(n = 7\)
\(d = \text{the common difference}\)
\(2 = 20 + (7-1)d\)
\(2 = 20 + 6d\)
\(2 - 20 = 6d\)
\(-18 = 6d\)
\(d = \dfrac{-18}{6}\)
\(d = -3\)
Answer: the common difference = \(-3\)
A later term, its term number, and the common difference are given.
\(u_n = u_1 + (n-1)d\)
\(u_{10} = 46\)
\(n = 10\)
\(d = 4\)
\(u_1 = \text{the first term}\)
\(46 = u_1 + (10-1) \times 4\)
\(46 = u_1 + 9 \times 4\)
\(46 = u_1 + 36\)
\(u_1 = 46 - 36\)
\(u_1 = 10\)
Answer: the first term = \(10\)
\(u_n = u_1 + (n-1)d\)
\(u_{12} = 49\)
\(n = 12\)
\(d = 4\)
\(u_1 = \text{the first term}\)
\(49 = u_1 + (12-1) \times 4\)
\(49 = u_1 + 11 \times 4\)
\(49 = u_1 + 44\)
\(u_1 = 49 - 44\)
\(u_1 = 5\)
Answer: the first term = \(5\)
\(u_n = u_1 + (n-1)d\)
\(u_{8} = 38\)
\(n = 8\)
\(d = 5\)
\(u_1 = \text{the first term}\)
\(38 = u_1 + (8-1) \times 5\)
\(38 = u_1 + 7 \times 5\)
\(38 = u_1 + 35\)
\(u_1 = 38 - 35\)
\(u_1 = 3\)
Answer: the first term = \(3\)
\(u_n = u_1 + (n-1)d\)
\(u_{15} = 72\)
\(n = 15\)
\(d = 3\)
\(u_1 = \text{the first term}\)
\(72 = u_1 + (15-1) \times 3\)
\(72 = u_1 + 14 \times 3\)
\(72 = u_1 + 42\)
\(u_1 = 72 - 42\)
\(u_1 = 30\)
Answer: the first term = \(30\)
\(u_n = u_1 + (n-1)d\)
\(u_{11} = 61\)
\(n = 11\)
\(d = 5\)
\(u_1 = \text{the first term}\)
\(61 = u_1 + (11-1) \times 5\)
\(61 = u_1 + 10 \times 5\)
\(61 = u_1 + 50\)
\(u_1 = 61 - 50\)
\(u_1 = 11\)
Answer: the first term = \(11\)
\(u_n = u_1 + (n-1)d\)
\(u_{9} = 4\)
\(n = 9\)
\(d = -2\)
\(u_1 = \text{the first term}\)
\(4 = u_1 + (9-1) \times \left(-2\right)\)
\(4 = u_1 + 8 \times \left(-2\right)\)
\(4 = u_1 - 16\)
\(u_1 = 4 + 16\)
\(u_1 = 20\)
Answer: the first term = \(20\)
The first term, the common difference, and the last term are given.
\(u_n = u_1 + (n-1)d\)
\(u_n = 46\)
\(u_1 = 6\)
\(d = 4\)
\(n = \text{the number of terms}\)
\(46 = 6 + (n-1) \times 4\)
\(46 - 6 = (n-1) \times 4\)
\(40 = (n-1) \times 4\)
\(n-1 = \dfrac{40}{4}\)
\(n-1 = 10\)
\(n = 11\)
Answer: 11 terms
\(u_n = u_1 + (n-1)d\)
\(u_n = 43\)
\(u_1 = 3\)
\(d = 5\)
\(n = \text{the number of terms}\)
\(43 = 3 + (n-1) \times 5\)
\(43 - 3 = (n-1) \times 5\)
\(40 = (n-1) \times 5\)
\(n-1 = \dfrac{40}{5}\)
\(n-1 = 8\)
\(n = 9\)
Answer: 9 terms
\(u_n = u_1 + (n-1)d\)
\(u_n = 62\)
\(u_1 = 8\)
\(d = 6\)
\(n = \text{the number of terms}\)
\(62 = 8 + (n-1) \times 6\)
\(62 - 8 = (n-1) \times 6\)
\(54 = (n-1) \times 6\)
\(n-1 = \dfrac{54}{6}\)
\(n-1 = 9\)
\(n = 10\)
Answer: 10 terms
\(u_n = u_1 + (n-1)d\)
\(u_n = -7\)
\(u_1 = 20\)
\(d = -3\)
\(n = \text{the number of terms}\)
\(-7 = 20 + (n-1) \times \left(-3\right)\)
\(-7 - 20 = (n-1) \times \left(-3\right)\)
\(-27 = (n-1) \times \left(-3\right)\)
\(n-1 = \dfrac{-27}{-3}\)
\(n-1 = 9\)
\(n = 10\)
Answer: 10 terms
\(u_n = u_1 + (n-1)d\)
\(u_n = 71\)
\(u_1 = 1\)
\(d = 7\)
\(n = \text{the number of terms}\)
\(71 = 1 + (n-1) \times 7\)
\(71 - 1 = (n-1) \times 7\)
\(70 = (n-1) \times 7\)
\(n-1 = \dfrac{70}{7}\)
\(n-1 = 10\)
\(n = 11\)
Answer: 11 terms
\(u_n = u_1 + (n-1)d\)
\(u_n = 55\)
\(u_1 = 15\)
\(d = 4\)
\(n = \text{the number of terms}\)
\(55 = 15 + (n-1) \times 4\)
\(55 - 15 = (n-1) \times 4\)
\(40 = (n-1) \times 4\)
\(n-1 = \dfrac{40}{4}\)
\(n-1 = 10\)
\(n = 11\)
Answer: 11 terms
Find the common difference, then the nth term.
The second term is one common difference after the first term.
\(d = 10 - 6\)
\(d = 4\)
\(u_n = u_1 + (n-1)d\)
\(u_{12} = \text{the term to find}\)
\(u_1 = 6\)
\(n = 12\)
\(d = 4\)
\(u_{12} = 6 + (12-1) \times 4\)
\(u_{12} = 6 + 11 \times 4\)
\(u_{12} = 6 + 44\)
\(u_{12} = 50\)
Answer: the common difference = \(4\) and the 12th term = \(50\)
The second term is one common difference after the first term.
\(d = 13 - 9\)
\(d = 4\)
\(u_n = u_1 + (n-1)d\)
\(u_{8} = \text{the term to find}\)
\(u_1 = 9\)
\(n = 8\)
\(d = 4\)
\(u_{8} = 9 + (8-1) \times 4\)
\(u_{8} = 9 + 7 \times 4\)
\(u_{8} = 9 + 28\)
\(u_{8} = 37\)
Answer: the common difference = \(4\) and the 8th term = \(37\)
The second term is one common difference after the first term.
\(d = 11 - 4\)
\(d = 7\)
\(u_n = u_1 + (n-1)d\)
\(u_{10} = \text{the term to find}\)
\(u_1 = 4\)
\(n = 10\)
\(d = 7\)
\(u_{10} = 4 + (10-1) \times 7\)
\(u_{10} = 4 + 9 \times 7\)
\(u_{10} = 4 + 63\)
\(u_{10} = 67\)
Answer: the common difference = \(7\) and the 10th term = \(67\)
The second term is one common difference after the first term.
\(d = 17 - 20\)
\(d = -3\)
\(u_n = u_1 + (n-1)d\)
\(u_{15} = \text{the term to find}\)
\(u_1 = 20\)
\(n = 15\)
\(d = -3\)
\(u_{15} = 20 + (15-1) \times \left(-3\right)\)
\(u_{15} = 20 + 14 \times \left(-3\right)\)
\(u_{15} = 20 - 42\)
\(u_{15} = -22\)
Answer: the common difference = \(-3\) and the 15th term = \(-22\)
The second term is one common difference after the first term.
\(d = 14 - 8\)
\(d = 6\)
\(u_n = u_1 + (n-1)d\)
\(u_{20} = \text{the term to find}\)
\(u_1 = 8\)
\(n = 20\)
\(d = 6\)
\(u_{20} = 8 + (20-1) \times 6\)
\(u_{20} = 8 + 19 \times 6\)
\(u_{20} = 8 + 114\)
\(u_{20} = 122\)
Answer: the common difference = \(6\) and the 20th term = \(122\)
The second term is one common difference after the first term.
\(d = 9 - 2\)
\(d = 7\)
\(u_n = u_1 + (n-1)d\)
\(u_{14} = \text{the term to find}\)
\(u_1 = 2\)
\(n = 14\)
\(d = 7\)
\(u_{14} = 2 + (14-1) \times 7\)
\(u_{14} = 2 + 13 \times 7\)
\(u_{14} = 2 + 91\)
\(u_{14} = 93\)
Answer: the common difference = \(7\) and the 14th term = \(93\)
Find the first term, then the nth term.
\(u_n = u_1 + (n-1)d\)
\(u_{2} = 15\)
\(n = 2\)
\(d = 4\)
\(u_1 = \text{the first term}\)
\(15 = u_1 + (2-1) \times 4\)
\(15 = u_1 + 1 \times 4\)
\(15 = u_1 + 4\)
\(u_1 = 15 - 4\)
\(u_1 = 11\)
\(u_n = u_1 + (n-1)d\)
\(u_{9} = \text{the term to find}\)
\(u_1 = 11\)
\(n = 9\)
\(d = 4\)
\(u_{9} = 11 + (9-1) \times 4\)
\(u_{9} = 11 + 8 \times 4\)
\(u_{9} = 11 + 32\)
\(u_{9} = 43\)
Answer: the first term = \(11\) and the 9th term = \(43\)
\(u_n = u_1 + (n-1)d\)
\(u_{2} = 18\)
\(n = 2\)
\(d = 5\)
\(u_1 = \text{the first term}\)
\(18 = u_1 + (2-1) \times 5\)
\(18 = u_1 + 1 \times 5\)
\(18 = u_1 + 5\)
\(u_1 = 18 - 5\)
\(u_1 = 13\)
\(u_n = u_1 + (n-1)d\)
\(u_{10} = \text{the term to find}\)
\(u_1 = 13\)
\(n = 10\)
\(d = 5\)
\(u_{10} = 13 + (10-1) \times 5\)
\(u_{10} = 13 + 9 \times 5\)
\(u_{10} = 13 + 45\)
\(u_{10} = 58\)
Answer: the first term = \(13\) and the 10th term = \(58\)
\(u_n = u_1 + (n-1)d\)
\(u_{2} = 7\)
\(n = 2\)
\(d = 3\)
\(u_1 = \text{the first term}\)
\(7 = u_1 + (2-1) \times 3\)
\(7 = u_1 + 1 \times 3\)
\(7 = u_1 + 3\)
\(u_1 = 7 - 3\)
\(u_1 = 4\)
\(u_n = u_1 + (n-1)d\)
\(u_{12} = \text{the term to find}\)
\(u_1 = 4\)
\(n = 12\)
\(d = 3\)
\(u_{12} = 4 + (12-1) \times 3\)
\(u_{12} = 4 + 11 \times 3\)
\(u_{12} = 4 + 33\)
\(u_{12} = 37\)
Answer: the first term = \(4\) and the 12th term = \(37\)
\(u_n = u_1 + (n-1)d\)
\(u_{2} = 20\)
\(n = 2\)
\(d = -4\)
\(u_1 = \text{the first term}\)
\(20 = u_1 + (2-1) \times \left(-4\right)\)
\(20 = u_1 + 1 \times \left(-4\right)\)
\(20 = u_1 - 4\)
\(u_1 = 20 + 4\)
\(u_1 = 24\)
\(u_n = u_1 + (n-1)d\)
\(u_{8} = \text{the term to find}\)
\(u_1 = 24\)
\(n = 8\)
\(d = -4\)
\(u_{8} = 24 + (8-1) \times \left(-4\right)\)
\(u_{8} = 24 + 7 \times \left(-4\right)\)
\(u_{8} = 24 - 28\)
\(u_{8} = -4\)
Answer: the first term = \(24\) and the 8th term = \(-4\)
\(u_n = u_1 + (n-1)d\)
\(u_{2} = 11\)
\(n = 2\)
\(d = 6\)
\(u_1 = \text{the first term}\)
\(11 = u_1 + (2-1) \times 6\)
\(11 = u_1 + 1 \times 6\)
\(11 = u_1 + 6\)
\(u_1 = 11 - 6\)
\(u_1 = 5\)
\(u_n = u_1 + (n-1)d\)
\(u_{15} = \text{the term to find}\)
\(u_1 = 5\)
\(n = 15\)
\(d = 6\)
\(u_{15} = 5 + (15-1) \times 6\)
\(u_{15} = 5 + 14 \times 6\)
\(u_{15} = 5 + 84\)
\(u_{15} = 89\)
Answer: the first term = \(5\) and the 15th term = \(89\)
\(u_n = u_1 + (n-1)d\)
\(u_{2} = 9\)
\(n = 2\)
\(d = -2\)
\(u_1 = \text{the first term}\)
\(9 = u_1 + (2-1) \times \left(-2\right)\)
\(9 = u_1 + 1 \times \left(-2\right)\)
\(9 = u_1 - 2\)
\(u_1 = 9 + 2\)
\(u_1 = 11\)
\(u_n = u_1 + (n-1)d\)
\(u_{20} = \text{the term to find}\)
\(u_1 = 11\)
\(n = 20\)
\(d = -2\)
\(u_{20} = 11 + (20-1) \times \left(-2\right)\)
\(u_{20} = 11 + 19 \times \left(-2\right)\)
\(u_{20} = 11 - 38\)
\(u_{20} = -27\)
Answer: the first term = \(11\) and the 20th term = \(-27\)
The last term is a numbered term. Find the common difference.
\(u_n = u_1 + (n-1)d\)
\(u_{9} = 28\)
\(u_1 = 4\)
\(n = 9\)
\(d = \text{the common difference}\)
\(28 = 4 + (9-1)d\)
\(28 = 4 + 8d\)
\(28 - 4 = 8d\)
\(24 = 8d\)
\(d = \dfrac{24}{8}\)
\(d = 3\)
Answer: the common difference = \(3\)
\(u_n = u_1 + (n-1)d\)
\(u_{11} = 46\)
\(u_1 = 6\)
\(n = 11\)
\(d = \text{the common difference}\)
\(46 = 6 + (11-1)d\)
\(46 = 6 + 10d\)
\(46 - 6 = 10d\)
\(40 = 10d\)
\(d = \dfrac{40}{10}\)
\(d = 4\)
Answer: the common difference = \(4\)
\(u_n = u_1 + (n-1)d\)
\(u_{8} = 38\)
\(u_1 = 3\)
\(n = 8\)
\(d = \text{the common difference}\)
\(38 = 3 + (8-1)d\)
\(38 = 3 + 7d\)
\(38 - 3 = 7d\)
\(35 = 7d\)
\(d = \dfrac{35}{7}\)
\(d = 5\)
Answer: the common difference = \(5\)
\(u_n = u_1 + (n-1)d\)
\(u_{16} = 5\)
\(u_1 = 50\)
\(n = 16\)
\(d = \text{the common difference}\)
\(5 = 50 + (16-1)d\)
\(5 = 50 + 15d\)
\(5 - 50 = 15d\)
\(-45 = 15d\)
\(d = \dfrac{-45}{15}\)
\(d = -3\)
Answer: the common difference = \(-3\)
\(u_n = u_1 + (n-1)d\)
\(u_{7} = 42\)
\(u_1 = 12\)
\(n = 7\)
\(d = \text{the common difference}\)
\(42 = 12 + (7-1)d\)
\(42 = 12 + 6d\)
\(42 - 12 = 6d\)
\(30 = 6d\)
\(d = \dfrac{30}{6}\)
\(d = 5\)
Answer: the common difference = \(5\)
\(u_n = u_1 + (n-1)d\)
\(u_{21} = 69\)
\(u_1 = 9\)
\(n = 21\)
\(d = \text{the common difference}\)
\(69 = 9 + (21-1)d\)
\(69 = 9 + 20d\)
\(69 - 9 = 20d\)
\(60 = 20d\)
\(d = \dfrac{60}{20}\)
\(d = 3\)
Answer: the common difference = \(3\)
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