IB AI · Unit 2 · 2B Geometric Sequences and Series
When the two given terms are not the first term, write the master formula once for each term, then divide to cancel u₁.
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SL 1.3
\(u_n = u_1 r^{n-1}\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1},\ r \neq 1\)
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r},\ r \neq 1\)
Use the master formula twice
Each known term gets its own column. Name that term as uₙ, and name its term number as n.
\(u_n = u_1 r^{n-1}\)
After both columns are simplified, only u₁ and r are unknown.
Divide to cancel u₁
Write the equation with the smaller term underneath. Divide the first equation by the second equation.
Substitute the common ratio back into the equation with the smaller term to find u₁.
Two later terms. The common ratio is positive.
Term 3
The 3rd term is \(12\).
\(u_n = 12\)
\(n = 3\)
\(u_n = u_1 r^{n-1}\)
\(12 = u_1 r^{2}\)
Term 6
The 6th term is \(96\).
\(u_n = 96\)
\(n = 6\)
\(u_n = u_1 r^{n-1}\)
\(96 = u_1 r^{5}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{96}{12} = \dfrac{r^{5}}{r^{2}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(12 = u_1 \times 2^{2}\)
\(2^{2} = 4\)
\(12 = u_1 \times 4\)
\(u_1 = 3\)
Answer: the first term = \(3\) and the common ratio = \(2\)
Term 2
The 2nd term is \(6\).
\(u_n = 6\)
\(n = 2\)
\(u_n = u_1 r^{n-1}\)
\(6 = u_1 r^{1}\)
Term 5
The 5th term is \(48\).
\(u_n = 48\)
\(n = 5\)
\(u_n = u_1 r^{n-1}\)
\(48 = u_1 r^{4}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{48}{6} = \dfrac{r^{4}}{r^{1}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(6 = u_1 \times 2^{1}\)
\(2^{1} = 2\)
\(6 = u_1 \times 2\)
\(u_1 = 3\)
Answer: the first term = \(3\) and the common ratio = \(2\)
Term 4
The 4th term is \(16\).
\(u_n = 16\)
\(n = 4\)
\(u_n = u_1 r^{n-1}\)
\(16 = u_1 r^{3}\)
Term 7
The 7th term is \(128\).
\(u_n = 128\)
\(n = 7\)
\(u_n = u_1 r^{n-1}\)
\(128 = u_1 r^{6}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{128}{16} = \dfrac{r^{6}}{r^{3}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(16 = u_1 \times 2^{3}\)
\(2^{3} = 8\)
\(16 = u_1 \times 8\)
\(u_1 = 2\)
Answer: the first term = \(2\) and the common ratio = \(2\)
Term 2
The 2nd term is \(10\).
\(u_n = 10\)
\(n = 2\)
\(u_n = u_1 r^{n-1}\)
\(10 = u_1 r^{1}\)
Term 5
The 5th term is \(80\).
\(u_n = 80\)
\(n = 5\)
\(u_n = u_1 r^{n-1}\)
\(80 = u_1 r^{4}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{80}{10} = \dfrac{r^{4}}{r^{1}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(10 = u_1 \times 2^{1}\)
\(2^{1} = 2\)
\(10 = u_1 \times 2\)
\(u_1 = 5\)
Answer: the first term = \(5\) and the common ratio = \(2\)
Term 3
The 3rd term is \(20\).
\(u_n = 20\)
\(n = 3\)
\(u_n = u_1 r^{n-1}\)
\(20 = u_1 r^{2}\)
Term 6
The 6th term is \(160\).
\(u_n = 160\)
\(n = 6\)
\(u_n = u_1 r^{n-1}\)
\(160 = u_1 r^{5}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{160}{20} = \dfrac{r^{5}}{r^{2}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(20 = u_1 \times 2^{2}\)
\(2^{2} = 4\)
\(20 = u_1 \times 4\)
\(u_1 = 5\)
Answer: the first term = \(5\) and the common ratio = \(2\)
Term 2
The 2nd term is \(9\).
\(u_n = 9\)
\(n = 2\)
\(u_n = u_1 r^{n-1}\)
\(9 = u_1 r^{1}\)
Term 5
The 5th term is \(243\).
\(u_n = 243\)
\(n = 5\)
\(u_n = u_1 r^{n-1}\)
\(243 = u_1 r^{4}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{243}{9} = \dfrac{r^{4}}{r^{1}}\)
\(27 = r^{3}\)
\(r = 3\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(9 = u_1 \times 3^{1}\)
\(3^{1} = 3\)
\(9 = u_1 \times 3\)
\(u_1 = 3\)
Answer: the first term = \(3\) and the common ratio = \(3\)
Find u₁ and r first. Then use either sum formula, or the calculator hack.
Term 4
The 4th term is \(24\).
\(u_n = 24\)
\(n = 4\)
\(u_n = u_1 r^{n-1}\)
\(24 = u_1 r^{3}\)
Term 7
The 7th term is \(192\).
\(u_n = 192\)
\(n = 7\)
\(u_n = u_1 r^{n-1}\)
\(192 = u_1 r^{6}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{192}{24} = \dfrac{r^{6}}{r^{3}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(24 = u_1 \times 2^{3}\)
\(2^{3} = 8\)
\(24 = u_1 \times 8\)
\(u_1 = 3\)
u₁ and r are known. Pick one way to find the sum of the first 6 terms.
Formula 1
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{6} = \dfrac{3\left(2^{6} - 1\right)}{2 - 1}\)
\(2^{6} = 64\)
\(64 - 1 = 63\)
\(2 - 1 = 1\)
\(S_{6} = \dfrac{3 \times 63}{1}\)
\(S_{6} = 189\)
Formula 2
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\)
\(S_{6} = \dfrac{3\left(1 - 2^{6}\right)}{1 - 2}\)
\(S_{6} = 189\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 3 \times 2^{n-1}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(3\times 2^{(x-1)}\).
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 6.
7. Press ENTER.
The calculator shows \(189\).
Answer: the sum = \(189\)
Term 2
The 2nd term is \(6\).
\(u_n = 6\)
\(n = 2\)
\(u_n = u_1 r^{n-1}\)
\(6 = u_1 r^{1}\)
Term 5
The 5th term is \(48\).
\(u_n = 48\)
\(n = 5\)
\(u_n = u_1 r^{n-1}\)
\(48 = u_1 r^{4}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{48}{6} = \dfrac{r^{4}}{r^{1}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(6 = u_1 \times 2^{1}\)
\(2^{1} = 2\)
\(6 = u_1 \times 2\)
\(u_1 = 3\)
u₁ and r are known. Pick one way to find the sum of the first 6 terms.
Formula 1
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{6} = \dfrac{3\left(2^{6} - 1\right)}{2 - 1}\)
\(2^{6} = 64\)
\(64 - 1 = 63\)
\(2 - 1 = 1\)
\(S_{6} = \dfrac{3 \times 63}{1}\)
\(S_{6} = 189\)
Formula 2
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\)
\(S_{6} = \dfrac{3\left(1 - 2^{6}\right)}{1 - 2}\)
\(S_{6} = 189\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 3 \times 2^{n-1}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(3\times 2^{(x-1)}\).
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 6.
7. Press ENTER.
The calculator shows \(189\).
Answer: the sum = \(189\)
Term 4
The 4th term is \(16\).
\(u_n = 16\)
\(n = 4\)
\(u_n = u_1 r^{n-1}\)
\(16 = u_1 r^{3}\)
Term 7
The 7th term is \(128\).
\(u_n = 128\)
\(n = 7\)
\(u_n = u_1 r^{n-1}\)
\(128 = u_1 r^{6}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{128}{16} = \dfrac{r^{6}}{r^{3}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(16 = u_1 \times 2^{3}\)
\(2^{3} = 8\)
\(16 = u_1 \times 8\)
\(u_1 = 2\)
u₁ and r are known. Pick one way to find the sum of the first 5 terms.
Formula 1
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{5} = \dfrac{2\left(2^{5} - 1\right)}{2 - 1}\)
\(2^{5} = 32\)
\(32 - 1 = 31\)
\(2 - 1 = 1\)
\(S_{5} = \dfrac{2 \times 31}{1}\)
\(S_{5} = 62\)
Formula 2
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\)
\(S_{5} = \dfrac{2\left(1 - 2^{5}\right)}{1 - 2}\)
\(S_{5} = 62\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 2^n\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(2^x\).
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 5.
7. Press ENTER.
The calculator shows \(62\).
Answer: the sum = \(62\)
Term 2
The 2nd term is \(10\).
\(u_n = 10\)
\(n = 2\)
\(u_n = u_1 r^{n-1}\)
\(10 = u_1 r^{1}\)
Term 5
The 5th term is \(80\).
\(u_n = 80\)
\(n = 5\)
\(u_n = u_1 r^{n-1}\)
\(80 = u_1 r^{4}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{80}{10} = \dfrac{r^{4}}{r^{1}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(10 = u_1 \times 2^{1}\)
\(2^{1} = 2\)
\(10 = u_1 \times 2\)
\(u_1 = 5\)
u₁ and r are known. Pick one way to find the sum of the first 7 terms.
Formula 1
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{7} = \dfrac{5\left(2^{7} - 1\right)}{2 - 1}\)
\(2^{7} = 128\)
\(128 - 1 = 127\)
\(2 - 1 = 1\)
\(S_{7} = \dfrac{5 \times 127}{1}\)
\(S_{7} = 635\)
Formula 2
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\)
\(S_{7} = \dfrac{5\left(1 - 2^{7}\right)}{1 - 2}\)
\(S_{7} = 635\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 5 \times 2^{n-1}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(5\times 2^{(x-1)}\).
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 \(635\).
Answer: the sum = \(635\)
Term 3
The 3rd term is \(20\).
\(u_n = 20\)
\(n = 3\)
\(u_n = u_1 r^{n-1}\)
\(20 = u_1 r^{2}\)
Term 6
The 6th term is \(160\).
\(u_n = 160\)
\(n = 6\)
\(u_n = u_1 r^{n-1}\)
\(160 = u_1 r^{5}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{160}{20} = \dfrac{r^{5}}{r^{2}}\)
\(8 = r^{3}\)
\(r = 2\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(20 = u_1 \times 2^{2}\)
\(2^{2} = 4\)
\(20 = u_1 \times 4\)
\(u_1 = 5\)
u₁ and r are known. Pick one way to find the sum of the first 6 terms.
Formula 1
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{6} = \dfrac{5\left(2^{6} - 1\right)}{2 - 1}\)
\(2^{6} = 64\)
\(64 - 1 = 63\)
\(2 - 1 = 1\)
\(S_{6} = \dfrac{5 \times 63}{1}\)
\(S_{6} = 315\)
Formula 2
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\)
\(S_{6} = \dfrac{5\left(1 - 2^{6}\right)}{1 - 2}\)
\(S_{6} = 315\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 5 \times 2^{n-1}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(5\times 2^{(x-1)}\).
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 6.
7. Press ENTER.
The calculator shows \(315\).
Answer: the sum = \(315\)
Term 2
The 2nd term is \(9\).
\(u_n = 9\)
\(n = 2\)
\(u_n = u_1 r^{n-1}\)
\(9 = u_1 r^{1}\)
Term 5
The 5th term is \(243\).
\(u_n = 243\)
\(n = 5\)
\(u_n = u_1 r^{n-1}\)
\(243 = u_1 r^{4}\)
Put the equation with the smaller term underneath. Divide the first equation by the second. That cancels u₁.
\(\dfrac{243}{9} = \dfrac{r^{4}}{r^{1}}\)
\(27 = r^{3}\)
\(r = 3\)
Substitute that common ratio into the second equation, the one with the smaller term.
\(9 = u_1 \times 3^{1}\)
\(3^{1} = 3\)
\(9 = u_1 \times 3\)
\(u_1 = 3\)
u₁ and r are known. Pick one way to find the sum of the first 5 terms.
Formula 1
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{5} = \dfrac{3\left(3^{5} - 1\right)}{3 - 1}\)
\(3^{5} = 243\)
\(243 - 1 = 242\)
\(3 - 1 = 2\)
\(S_{5} = \dfrac{3 \times 242}{2}\)
\(3 \times 242 = 726\)
\(S_{5} = \dfrac{726}{2}\)
\(S_{5} = 726 \div 2\)
\(S_{5} = 363\)
Formula 2
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\)
\(S_{5} = \dfrac{3\left(1 - 3^{5}\right)}{1 - 3}\)
\(S_{5} = 363\)
Calculator hack
Write the nth-term formula, then use summation.
\(u_n = 3^n\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(3^x\).
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 5.
7. Press ENTER.
The calculator shows \(363\).
Answer: the sum = \(363\)
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