IB AI · Unit 2 · 2B Geometric Sequences and Series
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 1.3
\(u_n = u_1 r^{n-1}\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1},\ r \neq 1\)
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r},\ r \neq 1\)
The nth term
\(u_n = u_1 r^{n-1}\)
\(u_n\) = last term
\(u_1\) = first term
\(n\) = number of terms
\(r\) = common ratio
Example. \(u_1 = 5\), \(r = 2\). Find \(u_4\).
\(u_4\) = the term to find
\(u_1 = 5\)
\(n = 4\)
\(r = 2\)
\(u_4 = 5 \times 2^{4-1}\)
\(u_4 = 5 \times 2^{3}\)
\(2^{3} = 8\)
\(u_4 = 5 \times 8\)
\(u_4 = 40\)
The first term and the common ratio are given.
\(u_n = u_1 r^{n-1}\)
\(u_{5} = \text{the term to find}\)
\(u_1 = 3\)
\(n = 5\)
\(r = 2\)
\(u_{5} = 3 \times 2^{4}\)
\(2^{4} = 16\)
\(u_{5} = 3 \times 16\)
\(u_{5} = 48\)
Answer: the 5th term = \(48\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = \text{the term to find}\)
\(u_1 = 4\)
\(n = 5\)
\(r = 3\)
\(u_{5} = 4 \times 3^{4}\)
\(3^{4} = 81\)
\(u_{5} = 4 \times 81\)
\(u_{5} = 324\)
Answer: the 5th term = \(324\)
\(u_n = u_1 r^{n-1}\)
\(u_{6} = \text{the term to find}\)
\(u_1 = 6\)
\(n = 6\)
\(r = 2\)
\(u_{6} = 6 \times 2^{5}\)
\(2^{5} = 32\)
\(u_{6} = 6 \times 32\)
\(u_{6} = 192\)
Answer: the 6th term = \(192\)
\(u_n = u_1 r^{n-1}\)
\(u_{7} = \text{the term to find}\)
\(u_1 = 5\)
\(n = 7\)
\(r = 2\)
\(u_{7} = 5 \times 2^{6}\)
\(2^{6} = 64\)
\(u_{7} = 5 \times 64\)
\(u_{7} = 320\)
Answer: the 7th term = \(320\)
\(u_n = u_1 r^{n-1}\)
\(u_{4} = \text{the term to find}\)
\(u_1 = 7\)
\(n = 4\)
\(r = 3\)
\(u_{4} = 7 \times 3^{3}\)
\(3^{3} = 27\)
\(u_{4} = 7 \times 27\)
\(u_{4} = 189\)
Answer: the 4th term = \(189\)
\(u_n = u_1 r^{n-1}\)
\(u_{8} = \text{the term to find}\)
\(u_1 = 10\)
\(n = 8\)
\(r = 2\)
\(u_{8} = 10 \times 2^{7}\)
\(2^{7} = 128\)
\(u_{8} = 10 \times 128\)
\(u_{8} = 1280\)
Answer: the 8th term = \(1280\)
The first term, the term number, and that term are given.
\(u_n = u_1 r^{n-1}\)
\(u_{4} = 81\)
\(u_1 = 3\)
\(n = 4\)
\(r = \text{the common ratio}\)
\(81 = 3 \times r^{3}\)
\(r^{3} = \dfrac{81}{3}\)
\(r^{3} = 27\)
\(3^{3} = 27\)
\(r = 3\)
Answer: the common ratio = \(3\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = 162\)
\(u_1 = 2\)
\(n = 5\)
\(r = \text{the common ratio}\)
\(162 = 2 \times r^{4}\)
\(r^{4} = \dfrac{162}{2}\)
\(r^{4} = 81\)
\(3^{4} = 81\)
\(r = 3\)
Answer: the common ratio = \(3\)
\(u_n = u_1 r^{n-1}\)
\(u_{6} = 128\)
\(u_1 = 4\)
\(n = 6\)
\(r = \text{the common ratio}\)
\(128 = 4 \times r^{5}\)
\(r^{5} = \dfrac{128}{4}\)
\(r^{5} = 32\)
\(2^{5} = 32\)
\(r = 2\)
Answer: the common ratio = \(2\)
\(u_n = u_1 r^{n-1}\)
\(u_{4} = 40\)
\(u_1 = 5\)
\(n = 4\)
\(r = \text{the common ratio}\)
\(40 = 5 \times r^{3}\)
\(r^{3} = \dfrac{40}{5}\)
\(r^{3} = 8\)
\(2^{3} = 8\)
\(r = 2\)
Answer: the common ratio = \(2\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = 96\)
\(u_1 = 6\)
\(n = 5\)
\(r = \text{the common ratio}\)
\(96 = 6 \times r^{4}\)
\(r^{4} = \dfrac{96}{6}\)
\(r^{4} = 16\)
\(2^{4} = 16\)
\(r = 2\)
Answer: the common ratio = \(2\)
\(u_n = u_1 r^{n-1}\)
\(u_{6} = 486\)
\(u_1 = 2\)
\(n = 6\)
\(r = \text{the common ratio}\)
\(486 = 2 \times r^{5}\)
\(r^{5} = \dfrac{486}{2}\)
\(r^{5} = 243\)
\(3^{5} = 243\)
\(r = 3\)
Answer: the common ratio = \(3\)
A later term, its term number, and the common ratio are given.
\(u_n = u_1 r^{n-1}\)
\(u_{6} = 96\)
\(n = 6\)
\(r = 2\)
\(u_1 = \text{the first term}\)
\(96 = u_1 \times 2^{5}\)
\(2^{5} = 32\)
\(96 = u_1 \times 32\)
\(u_1 = \dfrac{96}{32}\)
\(u_1 = 3\)
Answer: the first term = \(3\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = 48\)
\(n = 5\)
\(r = 2\)
\(u_1 = \text{the first term}\)
\(48 = u_1 \times 2^{4}\)
\(2^{4} = 16\)
\(48 = u_1 \times 16\)
\(u_1 = \dfrac{48}{16}\)
\(u_1 = 3\)
Answer: the first term = \(3\)
\(u_n = u_1 r^{n-1}\)
\(u_{4} = 54\)
\(n = 4\)
\(r = 3\)
\(u_1 = \text{the first term}\)
\(54 = u_1 \times 3^{3}\)
\(3^{3} = 27\)
\(54 = u_1 \times 27\)
\(u_1 = \dfrac{54}{27}\)
\(u_1 = 2\)
Answer: the first term = \(2\)
\(u_n = u_1 r^{n-1}\)
\(u_{6} = 160\)
\(n = 6\)
\(r = 2\)
\(u_1 = \text{the first term}\)
\(160 = u_1 \times 2^{5}\)
\(2^{5} = 32\)
\(160 = u_1 \times 32\)
\(u_1 = \dfrac{160}{32}\)
\(u_1 = 5\)
Answer: the first term = \(5\)
\(u_n = u_1 r^{n-1}\)
\(u_{4} = 250\)
\(n = 4\)
\(r = 5\)
\(u_1 = \text{the first term}\)
\(250 = u_1 \times 5^{3}\)
\(5^{3} = 125\)
\(250 = u_1 \times 125\)
\(u_1 = \dfrac{250}{125}\)
\(u_1 = 2\)
Answer: the first term = \(2\)
\(u_n = u_1 r^{n-1}\)
\(u_{7} = 192\)
\(n = 7\)
\(r = 2\)
\(u_1 = \text{the first term}\)
\(192 = u_1 \times 2^{6}\)
\(2^{6} = 64\)
\(192 = u_1 \times 64\)
\(u_1 = \dfrac{192}{64}\)
\(u_1 = 3\)
Answer: the first term = \(3\)
The first term, the common ratio, and the last term are given.
\(u_n = u_1 r^{n-1}\)
\(u_n = 48\)
\(u_1 = 3\)
\(r = 2\)
\(n = \text{the number of terms}\)
\(48 = 3 \times 2^{n-1}\)
\(2^{n-1} = \dfrac{48}{3}\)
\(2^{n-1} = 16\)
\(2^{4} = 16\)
\(n - 1 = 4\)
\(n = 5\)
Answer: 5 terms
\(u_n = u_1 r^{n-1}\)
\(u_n = 80\)
\(u_1 = 5\)
\(r = 2\)
\(n = \text{the number of terms}\)
\(80 = 5 \times 2^{n-1}\)
\(2^{n-1} = \dfrac{80}{5}\)
\(2^{n-1} = 16\)
\(2^{4} = 16\)
\(n - 1 = 4\)
\(n = 5\)
Answer: 5 terms
\(u_n = u_1 r^{n-1}\)
\(u_n = 96\)
\(u_1 = 3\)
\(r = 2\)
\(n = \text{the number of terms}\)
\(96 = 3 \times 2^{n-1}\)
\(2^{n-1} = \dfrac{96}{3}\)
\(2^{n-1} = 32\)
\(2^{5} = 32\)
\(n - 1 = 5\)
\(n = 6\)
Answer: 6 terms
\(u_n = u_1 r^{n-1}\)
\(u_n = 324\)
\(u_1 = 4\)
\(r = 3\)
\(n = \text{the number of terms}\)
\(324 = 4 \times 3^{n-1}\)
\(3^{n-1} = \dfrac{324}{4}\)
\(3^{n-1} = 81\)
\(3^{4} = 81\)
\(n - 1 = 4\)
\(n = 5\)
Answer: 5 terms
\(u_n = u_1 r^{n-1}\)
\(u_n = 224\)
\(u_1 = 7\)
\(r = 2\)
\(n = \text{the number of terms}\)
\(224 = 7 \times 2^{n-1}\)
\(2^{n-1} = \dfrac{224}{7}\)
\(2^{n-1} = 32\)
\(2^{5} = 32\)
\(n - 1 = 5\)
\(n = 6\)
Answer: 6 terms
\(u_n = u_1 r^{n-1}\)
\(u_n = 162\)
\(u_1 = 2\)
\(r = 3\)
\(n = \text{the number of terms}\)
\(162 = 2 \times 3^{n-1}\)
\(3^{n-1} = \dfrac{162}{2}\)
\(3^{n-1} = 81\)
\(3^{4} = 81\)
\(n - 1 = 4\)
\(n = 5\)
Answer: 5 terms
Find the common ratio, then the nth term.
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{6}{3}\)
\(r = 2\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = \text{the term to find}\)
\(u_1 = 3\)
\(n = 5\)
\(r = 2\)
\(u_{5} = 3 \times 2^{4}\)
\(2^{4} = 16\)
\(u_{5} = 3 \times 16\)
\(u_{5} = 48\)
Answer: the common ratio = \(2\) and the 5th term = \(48\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{12}{4}\)
\(r = 3\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = \text{the term to find}\)
\(u_1 = 4\)
\(n = 5\)
\(r = 3\)
\(u_{5} = 4 \times 3^{4}\)
\(3^{4} = 81\)
\(u_{5} = 4 \times 81\)
\(u_{5} = 324\)
Answer: the common ratio = \(3\) and the 5th term = \(324\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{10}{5}\)
\(r = 2\)
\(u_n = u_1 r^{n-1}\)
\(u_{6} = \text{the term to find}\)
\(u_1 = 5\)
\(n = 6\)
\(r = 2\)
\(u_{6} = 5 \times 2^{5}\)
\(2^{5} = 32\)
\(u_{6} = 5 \times 32\)
\(u_{6} = 160\)
Answer: the common ratio = \(2\) and the 6th term = \(160\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{18}{6}\)
\(r = 3\)
\(u_n = u_1 r^{n-1}\)
\(u_{4} = \text{the term to find}\)
\(u_1 = 6\)
\(n = 4\)
\(r = 3\)
\(u_{4} = 6 \times 3^{3}\)
\(3^{3} = 27\)
\(u_{4} = 6 \times 27\)
\(u_{4} = 162\)
Answer: the common ratio = \(3\) and the 4th term = \(162\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{16}{8}\)
\(r = 2\)
\(u_n = u_1 r^{n-1}\)
\(u_{8} = \text{the term to find}\)
\(u_1 = 8\)
\(n = 8\)
\(r = 2\)
\(u_{8} = 8 \times 2^{7}\)
\(2^{7} = 128\)
\(u_{8} = 8 \times 128\)
\(u_{8} = 1024\)
Answer: the common ratio = \(2\) and the 8th term = \(1024\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{14}{7}\)
\(r = 2\)
\(u_n = u_1 r^{n-1}\)
\(u_{7} = \text{the term to find}\)
\(u_1 = 7\)
\(n = 7\)
\(r = 2\)
\(u_{7} = 7 \times 2^{6}\)
\(2^{6} = 64\)
\(u_{7} = 7 \times 64\)
\(u_{7} = 448\)
Answer: the common ratio = \(2\) and the 7th term = \(448\)
Find the first term, then the nth term.
\(u_n = u_1 r^{n-1}\)
\(u_{2} = 20\)
\(n = 2\)
\(r = 2\)
\(u_1 = \text{the first term}\)
\(20 = u_1 \times 2^{1}\)
\(2^{1} = 2\)
\(20 = u_1 \times 2\)
\(u_1 = \dfrac{20}{2}\)
\(u_1 = 10\)
\(u_n = u_1 r^{n-1}\)
\(u_{6} = \text{the term to find}\)
\(u_1 = 10\)
\(n = 6\)
\(r = 2\)
\(u_{6} = 10 \times 2^{5}\)
\(2^{5} = 32\)
\(u_{6} = 10 \times 32\)
\(u_{6} = 320\)
Answer: the first term = \(10\) and the 6th term = \(320\)
\(u_n = u_1 r^{n-1}\)
\(u_{2} = 12\)
\(n = 2\)
\(r = 3\)
\(u_1 = \text{the first term}\)
\(12 = u_1 \times 3^{1}\)
\(3^{1} = 3\)
\(12 = u_1 \times 3\)
\(u_1 = \dfrac{12}{3}\)
\(u_1 = 4\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = \text{the term to find}\)
\(u_1 = 4\)
\(n = 5\)
\(r = 3\)
\(u_{5} = 4 \times 3^{4}\)
\(3^{4} = 81\)
\(u_{5} = 4 \times 81\)
\(u_{5} = 324\)
Answer: the first term = \(4\) and the 5th term = \(324\)
\(u_n = u_1 r^{n-1}\)
\(u_{2} = 10\)
\(n = 2\)
\(r = 2\)
\(u_1 = \text{the first term}\)
\(10 = u_1 \times 2^{1}\)
\(2^{1} = 2\)
\(10 = u_1 \times 2\)
\(u_1 = \dfrac{10}{2}\)
\(u_1 = 5\)
\(u_n = u_1 r^{n-1}\)
\(u_{6} = \text{the term to find}\)
\(u_1 = 5\)
\(n = 6\)
\(r = 2\)
\(u_{6} = 5 \times 2^{5}\)
\(2^{5} = 32\)
\(u_{6} = 5 \times 32\)
\(u_{6} = 160\)
Answer: the first term = \(5\) and the 6th term = \(160\)
\(u_n = u_1 r^{n-1}\)
\(u_{2} = 14\)
\(n = 2\)
\(r = 2\)
\(u_1 = \text{the first term}\)
\(14 = u_1 \times 2^{1}\)
\(2^{1} = 2\)
\(14 = u_1 \times 2\)
\(u_1 = \dfrac{14}{2}\)
\(u_1 = 7\)
\(u_n = u_1 r^{n-1}\)
\(u_{8} = \text{the term to find}\)
\(u_1 = 7\)
\(n = 8\)
\(r = 2\)
\(u_{8} = 7 \times 2^{7}\)
\(2^{7} = 128\)
\(u_{8} = 7 \times 128\)
\(u_{8} = 896\)
Answer: the first term = \(7\) and the 8th term = \(896\)
\(u_n = u_1 r^{n-1}\)
\(u_{2} = 18\)
\(n = 2\)
\(r = 3\)
\(u_1 = \text{the first term}\)
\(18 = u_1 \times 3^{1}\)
\(3^{1} = 3\)
\(18 = u_1 \times 3\)
\(u_1 = \dfrac{18}{3}\)
\(u_1 = 6\)
\(u_n = u_1 r^{n-1}\)
\(u_{4} = \text{the term to find}\)
\(u_1 = 6\)
\(n = 4\)
\(r = 3\)
\(u_{4} = 6 \times 3^{3}\)
\(3^{3} = 27\)
\(u_{4} = 6 \times 27\)
\(u_{4} = 162\)
Answer: the first term = \(6\) and the 4th term = \(162\)
\(u_n = u_1 r^{n-1}\)
\(u_{2} = 16\)
\(n = 2\)
\(r = 2\)
\(u_1 = \text{the first term}\)
\(16 = u_1 \times 2^{1}\)
\(2^{1} = 2\)
\(16 = u_1 \times 2\)
\(u_1 = \dfrac{16}{2}\)
\(u_1 = 8\)
\(u_n = u_1 r^{n-1}\)
\(u_{7} = \text{the term to find}\)
\(u_1 = 8\)
\(n = 7\)
\(r = 2\)
\(u_{7} = 8 \times 2^{6}\)
\(2^{6} = 64\)
\(u_{7} = 8 \times 64\)
\(u_{7} = 512\)
Answer: the first term = \(8\) and the 7th term = \(512\)
The last term is a numbered term. Find the common ratio.
\(u_n = u_1 r^{n-1}\)
\(u_{6} = 128\)
\(u_1 = 4\)
\(n = 6\)
\(r = \text{the common ratio}\)
\(128 = 4 \times r^{5}\)
\(r^{5} = \dfrac{128}{4}\)
\(r^{5} = 32\)
\(2^{5} = 32\)
\(r = 2\)
Answer: the common ratio = \(2\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = 162\)
\(u_1 = 2\)
\(n = 5\)
\(r = \text{the common ratio}\)
\(162 = 2 \times r^{4}\)
\(r^{4} = \dfrac{162}{2}\)
\(r^{4} = 81\)
\(3^{4} = 81\)
\(r = 3\)
Answer: the common ratio = \(3\)
\(u_n = u_1 r^{n-1}\)
\(u_{6} = 96\)
\(u_1 = 3\)
\(n = 6\)
\(r = \text{the common ratio}\)
\(96 = 3 \times r^{5}\)
\(r^{5} = \dfrac{96}{3}\)
\(r^{5} = 32\)
\(2^{5} = 32\)
\(r = 2\)
Answer: the common ratio = \(2\)
\(u_n = u_1 r^{n-1}\)
\(u_{4} = 108\)
\(u_1 = 4\)
\(n = 4\)
\(r = \text{the common ratio}\)
\(108 = 4 \times r^{3}\)
\(r^{3} = \dfrac{108}{4}\)
\(r^{3} = 27\)
\(3^{3} = 27\)
\(r = 3\)
Answer: the common ratio = \(3\)
\(u_n = u_1 r^{n-1}\)
\(u_{7} = 320\)
\(u_1 = 5\)
\(n = 7\)
\(r = \text{the common ratio}\)
\(320 = 5 \times r^{6}\)
\(r^{6} = \dfrac{320}{5}\)
\(r^{6} = 64\)
\(2^{6} = 64\)
\(r = 2\)
Answer: the common ratio = \(2\)
\(u_n = u_1 r^{n-1}\)
\(u_{5} = 96\)
\(u_1 = 6\)
\(n = 5\)
\(r = \text{the common ratio}\)
\(96 = 6 \times r^{4}\)
\(r^{4} = \dfrac{96}{6}\)
\(r^{4} = 16\)
\(2^{4} = 16\)
\(r = 2\)
Answer: the common ratio = \(2\)
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