IB AI · Unit 2 · 2A Arithmetic Sequences
When the two given terms are not the first term, write the master formula once for each term, then eliminate u₁.
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SL 1.2
\(u_n = u_1 + (n-1)d\)
\(S_n = \dfrac{n}{2}\bigl(2u_1 + (n-1)d\bigr)\)
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
Use the master formula twice
Each known term gets its own column. Name that term as uₙ, and name its term number as n.
\(u_n = u_1 + (n-1)d\)
After both columns are simplified, only u₁ and d are unknown.
Eliminate u₁
Write the equation with the smaller term underneath the other equation. Make each term of that second equation negative, then add.
Substitute the common difference back into the first equation to find u₁.
Two later terms. Find the first term and the common difference.
Term 4
The 4th term is \(18\).
\(u_n = 18\)
\(n = 4\)
\(u_n = u_1 + (n-1)d\)
\(18 = u_1 + (4-1)d\)
\(18 = u_1 + 3d\)
Term 10
The 10th term is \(48\).
\(u_n = 48\)
\(n = 10\)
\(u_n = u_1 + (n-1)d\)
\(48 = u_1 + (10-1)d\)
\(48 = u_1 + 9d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{30}{6}\)
\(d = 5\)
Substitute that common difference into the first equation.
\(48 = u_1 + 9 \times 5\)
\(48 = u_1 + 45\)
\(u_1 = 48 - 45\)
\(u_1 = 3\)
Answer: the first term = \(3\) and the common difference = \(5\)
Term 3
The 3rd term is \(11\).
\(u_n = 11\)
\(n = 3\)
\(u_n = u_1 + (n-1)d\)
\(11 = u_1 + (3-1)d\)
\(11 = u_1 + 2d\)
Term 8
The 8th term is \(31\).
\(u_n = 31\)
\(n = 8\)
\(u_n = u_1 + (n-1)d\)
\(31 = u_1 + (8-1)d\)
\(31 = u_1 + 7d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{20}{5}\)
\(d = 4\)
Substitute that common difference into the first equation.
\(31 = u_1 + 7 \times 4\)
\(31 = u_1 + 28\)
\(u_1 = 31 - 28\)
\(u_1 = 3\)
Answer: the first term = \(3\) and the common difference = \(4\)
Term 5
The 5th term is \(20\).
\(u_n = 20\)
\(n = 5\)
\(u_n = u_1 + (n-1)d\)
\(20 = u_1 + (5-1)d\)
\(20 = u_1 + 4d\)
Term 9
The 9th term is \(36\).
\(u_n = 36\)
\(n = 9\)
\(u_n = u_1 + (n-1)d\)
\(36 = u_1 + (9-1)d\)
\(36 = u_1 + 8d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{16}{4}\)
\(d = 4\)
Substitute that common difference into the first equation.
\(36 = u_1 + 8 \times 4\)
\(36 = u_1 + 32\)
\(u_1 = 36 - 32\)
\(u_1 = 4\)
Answer: the first term = \(4\) and the common difference = \(4\)
Term 2
The 2nd term is \(9\).
\(u_n = 9\)
\(n = 2\)
\(u_n = u_1 + (n-1)d\)
\(9 = u_1 + (2-1)d\)
\(9 = u_1 + 1d\)
Term 7
The 7th term is \(29\).
\(u_n = 29\)
\(n = 7\)
\(u_n = u_1 + (n-1)d\)
\(29 = u_1 + (7-1)d\)
\(29 = u_1 + 6d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{20}{5}\)
\(d = 4\)
Substitute that common difference into the first equation.
\(29 = u_1 + 6 \times 4\)
\(29 = u_1 + 24\)
\(u_1 = 29 - 24\)
\(u_1 = 5\)
Answer: the first term = \(5\) and the common difference = \(4\)
Term 6
The 6th term is \(28\).
\(u_n = 28\)
\(n = 6\)
\(u_n = u_1 + (n-1)d\)
\(28 = u_1 + (6-1)d\)
\(28 = u_1 + 5d\)
Term 11
The 11th term is \(53\).
\(u_n = 53\)
\(n = 11\)
\(u_n = u_1 + (n-1)d\)
\(53 = u_1 + (11-1)d\)
\(53 = u_1 + 10d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{25}{5}\)
\(d = 5\)
Substitute that common difference into the first equation.
\(53 = u_1 + 10 \times 5\)
\(53 = u_1 + 50\)
\(u_1 = 53 - 50\)
\(u_1 = 3\)
Answer: the first term = \(3\) and the common difference = \(5\)
Term 4
The 4th term is \(14\).
\(u_n = 14\)
\(n = 4\)
\(u_n = u_1 + (n-1)d\)
\(14 = u_1 + (4-1)d\)
\(14 = u_1 + 3d\)
Term 12
The 12th term is \(38\).
\(u_n = 38\)
\(n = 12\)
\(u_n = u_1 + (n-1)d\)
\(38 = u_1 + (12-1)d\)
\(38 = u_1 + 11d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{24}{8}\)
\(d = 3\)
Substitute that common difference into the first equation.
\(38 = u_1 + 11 \times 3\)
\(38 = u_1 + 33\)
\(u_1 = 38 - 33\)
\(u_1 = 5\)
Answer: the first term = \(5\) and the common difference = \(3\)
Find u₁ and d first. Then use either sum formula, or the calculator hack.
Term 4
The 4th term is \(22\).
\(u_n = 22\)
\(n = 4\)
\(u_n = u_1 + (n-1)d\)
\(22 = u_1 + (4-1)d\)
\(22 = u_1 + 3d\)
Term 9
The 9th term is \(42\).
\(u_n = 42\)
\(n = 9\)
\(u_n = u_1 + (n-1)d\)
\(42 = u_1 + (9-1)d\)
\(42 = u_1 + 8d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{20}{5}\)
\(d = 4\)
Substitute that common difference into the first equation.
\(42 = u_1 + 8 \times 4\)
\(42 = u_1 + 32\)
\(u_1 = 42 - 32\)
\(u_1 = 10\)
u₁ and d are known. Pick one way to find the sum of the first 10 terms.
Formula 1
\(S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right)\)
\(S_{10} = \dfrac{10}{2}\left(2 \times 10 + (10-1) \times 4\right)\)
\(S_{10} = \dfrac{10}{2}\left(20 + 9 \times 4\right)\)
\(S_{10} = \dfrac{10}{2}\left(20 + 36\right)\)
\(S_{10} = \dfrac{10}{2} \times 56\)
\(S_{10} = 5 \times 56\)
\(S_{10} = 280\)
Formula 2
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{10} = \dfrac{10}{2}\left(10 + 46\right)\)
\(10 + 46 = 56\)
\(S_{10} = \dfrac{10}{2} \times 56\)
\(S_{10} = 5 \times 56\)
\(S_{10} = 280\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 4n + 6\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(4x + 6\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 10.
7. Press ENTER.
The calculator shows \(280\).
Answer: the sum = \(280\)
Term 3
The 3rd term is \(14\).
\(u_n = 14\)
\(n = 3\)
\(u_n = u_1 + (n-1)d\)
\(14 = u_1 + (3-1)d\)
\(14 = u_1 + 2d\)
Term 8
The 8th term is \(34\).
\(u_n = 34\)
\(n = 8\)
\(u_n = u_1 + (n-1)d\)
\(34 = u_1 + (8-1)d\)
\(34 = u_1 + 7d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{20}{5}\)
\(d = 4\)
Substitute that common difference into the first equation.
\(34 = u_1 + 7 \times 4\)
\(34 = u_1 + 28\)
\(u_1 = 34 - 28\)
\(u_1 = 6\)
u₁ and d are known. Pick one way to find the sum of the first 8 terms.
Formula 1
\(S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right)\)
\(S_{8} = \dfrac{8}{2}\left(2 \times 6 + (8-1) \times 4\right)\)
\(S_{8} = \dfrac{8}{2}\left(12 + 7 \times 4\right)\)
\(S_{8} = \dfrac{8}{2}\left(12 + 28\right)\)
\(S_{8} = \dfrac{8}{2} \times 40\)
\(S_{8} = 4 \times 40\)
\(S_{8} = 160\)
Formula 2
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{8} = \dfrac{8}{2}\left(6 + 34\right)\)
\(6 + 34 = 40\)
\(S_{8} = \dfrac{8}{2} \times 40\)
\(S_{8} = 4 \times 40\)
\(S_{8} = 160\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 4n + 2\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(4x + 2\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 8.
7. Press ENTER.
The calculator shows \(160\).
Answer: the sum = \(160\)
Term 5
The 5th term is \(17\).
\(u_n = 17\)
\(n = 5\)
\(u_n = u_1 + (n-1)d\)
\(17 = u_1 + (5-1)d\)
\(17 = u_1 + 4d\)
Term 9
The 9th term is \(33\).
\(u_n = 33\)
\(n = 9\)
\(u_n = u_1 + (n-1)d\)
\(33 = u_1 + (9-1)d\)
\(33 = u_1 + 8d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{16}{4}\)
\(d = 4\)
Substitute that common difference into the first equation.
\(33 = u_1 + 8 \times 4\)
\(33 = u_1 + 32\)
\(u_1 = 33 - 32\)
\(u_1 = 1\)
u₁ and d are known. Pick one way to find the sum of the first 12 terms.
Formula 1
\(S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right)\)
\(S_{12} = \dfrac{12}{2}\left(2 \times 1 + (12-1) \times 4\right)\)
\(S_{12} = \dfrac{12}{2}\left(2 + 11 \times 4\right)\)
\(S_{12} = \dfrac{12}{2}\left(2 + 44\right)\)
\(S_{12} = \dfrac{12}{2} \times 46\)
\(S_{12} = 6 \times 46\)
\(S_{12} = 276\)
Formula 2
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{12} = \dfrac{12}{2}\left(1 + 45\right)\)
\(1 + 45 = 46\)
\(S_{12} = \dfrac{12}{2} \times 46\)
\(S_{12} = 6 \times 46\)
\(S_{12} = 276\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 4n - 3\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(4x - 3\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 12.
7. Press ENTER.
The calculator shows \(276\).
Answer: the sum = \(276\)
Term 2
The 2nd term is \(13\).
\(u_n = 13\)
\(n = 2\)
\(u_n = u_1 + (n-1)d\)
\(13 = u_1 + (2-1)d\)
\(13 = u_1 + 1d\)
Term 6
The 6th term is \(33\).
\(u_n = 33\)
\(n = 6\)
\(u_n = u_1 + (n-1)d\)
\(33 = u_1 + (6-1)d\)
\(33 = u_1 + 5d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{20}{4}\)
\(d = 5\)
Substitute that common difference into the first equation.
\(33 = u_1 + 5 \times 5\)
\(33 = u_1 + 25\)
\(u_1 = 33 - 25\)
\(u_1 = 8\)
u₁ and d are known. Pick one way to find the sum of the first 7 terms.
Formula 1
\(S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right)\)
\(S_{7} = \dfrac{7}{2}\left(2 \times 8 + (7-1) \times 5\right)\)
\(S_{7} = \dfrac{7}{2}\left(16 + 6 \times 5\right)\)
\(S_{7} = \dfrac{7}{2}\left(16 + 30\right)\)
\(S_{7} = \dfrac{7}{2} \times 46\)
\(S_{7} = 7 \times 23\)
\(S_{7} = 161\)
Formula 2
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{7} = \dfrac{7}{2}\left(8 + 38\right)\)
\(8 + 38 = 46\)
\(S_{7} = \dfrac{7}{2} \times 46\)
\(S_{7} = 7 \times 23\)
\(S_{7} = 161\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 5n + 3\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(5x + 3\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 7.
7. Press ENTER.
The calculator shows \(161\).
Answer: the sum = \(161\)
Term 4
The 4th term is \(19\).
\(u_n = 19\)
\(n = 4\)
\(u_n = u_1 + (n-1)d\)
\(19 = u_1 + (4-1)d\)
\(19 = u_1 + 3d\)
Term 10
The 10th term is \(43\).
\(u_n = 43\)
\(n = 10\)
\(u_n = u_1 + (n-1)d\)
\(43 = u_1 + (10-1)d\)
\(43 = u_1 + 9d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{24}{6}\)
\(d = 4\)
Substitute that common difference into the first equation.
\(43 = u_1 + 9 \times 4\)
\(43 = u_1 + 36\)
\(u_1 = 43 - 36\)
\(u_1 = 7\)
u₁ and d are known. Pick one way to find the sum of the first 11 terms.
Formula 1
\(S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right)\)
\(S_{11} = \dfrac{11}{2}\left(2 \times 7 + (11-1) \times 4\right)\)
\(S_{11} = \dfrac{11}{2}\left(14 + 10 \times 4\right)\)
\(S_{11} = \dfrac{11}{2}\left(14 + 40\right)\)
\(S_{11} = \dfrac{11}{2} \times 54\)
\(S_{11} = 11 \times 27\)
\(S_{11} = 297\)
Formula 2
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{11} = \dfrac{11}{2}\left(7 + 47\right)\)
\(7 + 47 = 54\)
\(S_{11} = \dfrac{11}{2} \times 54\)
\(S_{11} = 11 \times 27\)
\(S_{11} = 297\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 4n + 3\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(4x + 3\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 11.
7. Press ENTER.
The calculator shows \(297\).
Answer: the sum = \(297\)
Term 6
The 6th term is \(26\).
\(u_n = 26\)
\(n = 6\)
\(u_n = u_1 + (n-1)d\)
\(26 = u_1 + (6-1)d\)
\(26 = u_1 + 5d\)
Term 15
The 15th term is \(62\).
\(u_n = 62\)
\(n = 15\)
\(u_n = u_1 + (n-1)d\)
\(62 = u_1 + (15-1)d\)
\(62 = u_1 + 14d\)
Put the equation with the smaller term underneath, and make each term of that equation negative.
\(d = \dfrac{36}{9}\)
\(d = 4\)
Substitute that common difference into the first equation.
\(62 = u_1 + 14 \times 4\)
\(62 = u_1 + 56\)
\(u_1 = 62 - 56\)
\(u_1 = 6\)
u₁ and d are known. Pick one way to find the sum of the first 15 terms.
Formula 1
\(S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right)\)
\(S_{15} = \dfrac{15}{2}\left(2 \times 6 + (15-1) \times 4\right)\)
\(S_{15} = \dfrac{15}{2}\left(12 + 14 \times 4\right)\)
\(S_{15} = \dfrac{15}{2}\left(12 + 56\right)\)
\(S_{15} = \dfrac{15}{2} \times 68\)
\(S_{15} = 15 \times 34\)
\(S_{15} = 510\)
Formula 2
\(S_n = \dfrac{n}{2}(u_1 + u_n)\)
\(S_{15} = \dfrac{15}{2}\left(6 + 62\right)\)
\(6 + 62 = 68\)
\(S_{15} = \dfrac{15}{2} \times 68\)
\(S_{15} = 15 \times 34\)
\(S_{15} = 510\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 4n + 2\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(4x + 2\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 15.
7. Press ENTER.
The calculator shows \(510\).
Answer: the sum = \(510\)
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