IB AI · Unit 2 · 2B Geometric Sequences and Series
Substitute the first term and the common ratio, then simplify. For algebraic terms, divide to find r, check the third term, and simplify the powers.
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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 formula
\(u_n = u_1 r^{n-1}\)
\(u_n\) = the formula
\(u_1\) = first term
\(n\) = the term number
\(r\) = common ratio
Example. \(u_1 = 6\), \(r = 2\). Write a formula for \(u_n\).
\(u_1 = 6\)
\(r = 2\)
\(u_n = 6 \times 2^{n-1}\)
\(u_n = 3 \times 2 \times 2^{n-1}\)
\(u_n = 3 \times 2^{n}\)
Algebraic terms
The common ratio is the second term divided by the first term.
\(r = \dfrac{u_2}{u_1}\)
Check it on the third term, then simplify \(u_n = u_1 r^{n-1}\).
Substitute, then simplify when a factor of r divides the first term.
\(u_n = u_1 r^{n-1}\)
\(u_1 = 8\)
\(r = 2\)
\(u_n = 8 \times 2^{n-1}\)
\(u_n = 4 \times 2 \times 2^{n-1}\)
\(u_n = 4 \times 2^n\)
Answer: \(u_n = 4 \times 2^n\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 6\)
\(r = 2\)
\(u_n = 6 \times 2^{n-1}\)
\(u_n = 3 \times 2 \times 2^{n-1}\)
\(u_n = 3 \times 2^n\)
Answer: \(u_n = 3 \times 2^n\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 5\)
\(r = 3\)
\(u_n = 5 \times 3^{n-1}\)
Answer: \(u_n = 5 \times 3^{n-1}\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 12\)
\(r = 2\)
\(u_n = 12 \times 2^{n-1}\)
\(u_n = 6 \times 2 \times 2^{n-1}\)
\(u_n = 6 \times 2^n\)
Answer: \(u_n = 6 \times 2^n\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 9\)
\(r = 3\)
\(u_n = 9 \times 3^{n-1}\)
\(u_n = 3 \times 3 \times 3^{n-1}\)
\(u_n = 3 \times 3^n\)
Answer: \(u_n = 3 \times 3^n\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 4\)
\(r = 5\)
\(u_n = 4 \times 5^{n-1}\)
Answer: \(u_n = 4 \times 5^{n-1}\)
Find the common ratio, then the formula.
\(r = \dfrac{12}{6}\)
\(r = 2\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 6\)
\(r = 2\)
\(u_n = 6 \times 2^{n-1}\)
\(u_n = 3 \times 2 \times 2^{n-1}\)
\(u_n = 3 \times 2^n\)
Answer: \(u_n = 3 \times 2^n\)
\(r = \dfrac{10}{5}\)
\(r = 2\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 5\)
\(r = 2\)
\(u_n = 5 \times 2^{n-1}\)
Answer: \(u_n = 5 \times 2^{n-1}\)
\(r = \dfrac{12}{4}\)
\(r = 3\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 4\)
\(r = 3\)
\(u_n = 4 \times 3^{n-1}\)
Answer: \(u_n = 4 \times 3^{n-1}\)
\(r = \dfrac{14}{7}\)
\(r = 2\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 7\)
\(r = 2\)
\(u_n = 7 \times 2^{n-1}\)
Answer: \(u_n = 7 \times 2^{n-1}\)
\(r = \dfrac{9}{3}\)
\(r = 3\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 3\)
\(r = 3\)
\(u_n = 3 \times 3^{n-1}\)
\(u_n = 1 \times 3 \times 3^{n-1}\)
\(u_n = 3^n\)
Answer: \(u_n = 3^n\)
\(r = \dfrac{16}{8}\)
\(r = 2\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 8\)
\(r = 2\)
\(u_n = 8 \times 2^{n-1}\)
\(u_n = 4 \times 2 \times 2^{n-1}\)
\(u_n = 4 \times 2^n\)
Answer: \(u_n = 4 \times 2^n\)
Read off the first term and the common ratio, then write the formula.
\(u_n = u_1 r^{n-1}\)
\(u_1 = 7\)
\(r = 2\)
\(u_n = 7 \times 2^{n-1}\)
Answer: \(u_n = 7 \times 2^{n-1}\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 5\)
\(r = 3\)
\(u_n = 5 \times 3^{n-1}\)
Answer: \(u_n = 5 \times 3^{n-1}\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 12\)
\(r = 2\)
\(u_n = 12 \times 2^{n-1}\)
\(u_n = 6 \times 2 \times 2^{n-1}\)
\(u_n = 6 \times 2^n\)
Answer: \(u_n = 6 \times 2^n\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 9\)
\(r = 3\)
\(u_n = 9 \times 3^{n-1}\)
\(u_n = 3 \times 3 \times 3^{n-1}\)
\(u_n = 3 \times 3^n\)
Answer: \(u_n = 3 \times 3^n\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 4\)
\(r = 2\)
\(u_n = 4 \times 2^{n-1}\)
\(u_n = 2 \times 2 \times 2^{n-1}\)
\(u_n = 2 \times 2^n\)
Answer: \(u_n = 2 \times 2^n\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 10\)
\(r = 2\)
\(u_n = 10 \times 2^{n-1}\)
\(u_n = 5 \times 2 \times 2^{n-1}\)
\(u_n = 5 \times 2^n\)
Answer: \(u_n = 5 \times 2^n\)
Divide, check the third term, then simplify the powers.
Divide the second term by the first term.
\(r = \dfrac{6x^{2}}{2x}\)
\(r = \dfrac{6}{2}x^{2-1}\)
\(r = 3x\)
Multiply the second term by r. The product is the third term.
\(6x^{2} \times 3x = 18x^{3}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = 2x \times \left(3x\right)^{n-1}\)
\(u_n = 2x \cdot 3^{n-1}x^{n-1}\)
\(u_n = 2 \cdot 3^{n-1}x^{1+(n-1)}\)
\(u_n = 2 \cdot 3^{n-1}x^{n}\)
Answer: the nth term \(u_n = 2 \cdot 3^{n-1}x^{n}\)
Divide the second term by the first term.
\(r = \dfrac{10a}{5}\)
\(r = \dfrac{10}{5}a^{1-0}\)
\(r = 2a\)
Multiply the second term by r. The product is the third term.
\(10a \times 2a = 20a^{2}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = 5 \times \left(2a\right)^{n-1}\)
\(u_n = 5 \cdot 2^{n-1}a^{n-1}\)
Answer: the nth term \(u_n = 5 \cdot 2^{n-1}a^{n-1}\)
Divide the second term by the first term.
\(r = \dfrac{a^{5}}{a^{2}}\)
\(r = a^{5-2}\)
\(r = a^{3}\)
Multiply the second term by r. The product is the third term.
\(a^{5} \times a^{3} = a^{8}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = a^{2} \times \left(a^{3}\right)^{n-1}\)
\(u_n = a^{2}a^{3(n-1)}\)
\(u_n = a^{2+3(n-1)}\)
\(u_n = a^{3n-1}\)
Answer: the nth term \(u_n = a^{3n-1}\)
Divide the second term by the first term.
\(r = \dfrac{12y^{3}}{4y}\)
\(r = \dfrac{12}{4}y^{3-1}\)
\(r = 3y^{2}\)
Multiply the second term by r. The product is the third term.
\(12y^{3} \times 3y^{2} = 36y^{5}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = 4y \times \left(3y^{2}\right)^{n-1}\)
\(u_n = 4y \cdot 3^{n-1}y^{2(n-1)}\)
\(u_n = 4 \cdot 3^{n-1}y^{1+2(n-1)}\)
\(u_n = 4 \cdot 3^{n-1}y^{2n-1}\)
Answer: the nth term \(u_n = 4 \cdot 3^{n-1}y^{2n-1}\)
Divide the second term by the first term.
\(r = \dfrac{6b^{3}}{3b^{2}}\)
\(r = \dfrac{6}{3}b^{3-2}\)
\(r = 2b\)
Multiply the second term by r. The product is the third term.
\(6b^{3} \times 2b = 12b^{4}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = 3b^{2} \times \left(2b\right)^{n-1}\)
\(u_n = 3b^{2} \cdot 2^{n-1}b^{n-1}\)
\(u_n = 3 \cdot 2^{n-1}b^{2+(n-1)}\)
\(u_n = 3 \cdot 2^{n-1}b^{n+1}\)
Answer: the nth term \(u_n = 3 \cdot 2^{n-1}b^{n+1}\)
Divide the second term by the first term.
\(r = \dfrac{7k^{2}}{7k}\)
\(r = k^{2-1}\)
\(r = k\)
Multiply the second term by r. The product is the third term.
\(7k^{2} \times k = 7k^{3}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = 7k \times k^{n-1}\)
\(u_n = 7kk^{n-1}\)
\(u_n = 7k^{1+(n-1)}\)
\(u_n = 7k^{n}\)
Answer: the nth term \(u_n = 7k^{n}\)
The ratio is often a quotient, so some powers in later terms are negative.
Divide the second term by the first term.
\(r = \dfrac{xy^{2}}{x^{2}y}\)
\(r = x^{1-2}y^{2-1}\)
\(r = \dfrac{y}{x}\)
Multiply the second term by r. The product is the third term.
\(xy^{2} \times \left(\dfrac{y}{x}\right) = y^{3}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = x^{2}y \times \left(\dfrac{y}{x}\right)^{n-1}\)
\(u_n = x^{2}y \cdot \dfrac{y^{n-1}}{x^{n-1}}\)
\(u_n = x^{2-(n-1)}y^{1+(n-1)}\)
\(u_n = x^{3-n}y^{n}\)
Answer: the nth term \(u_n = x^{3-n}y^{n}\)
Divide the second term by the first term.
\(r = \dfrac{a^{2}b^{2}}{a^{3}b}\)
\(r = a^{2-3}b^{2-1}\)
\(r = \dfrac{b}{a}\)
Multiply the second term by r. The product is the third term.
\(a^{2}b^{2} \times \left(\dfrac{b}{a}\right) = ab^{3}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = a^{3}b \times \left(\dfrac{b}{a}\right)^{n-1}\)
\(u_n = a^{3}b \cdot \dfrac{b^{n-1}}{a^{n-1}}\)
\(u_n = a^{3-(n-1)}b^{1+(n-1)}\)
\(u_n = a^{4-n}b^{n}\)
Answer: the nth term \(u_n = a^{4-n}b^{n}\)
Divide the second term by the first term.
\(r = \dfrac{mn^{4}}{m^{2}n^{3}}\)
\(r = m^{1-2}n^{4-3}\)
\(r = \dfrac{n}{m}\)
Multiply the second term by r. The product is the third term.
\(mn^{4} \times \left(\dfrac{n}{m}\right) = n^{5}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = m^{2}n^{3} \times \left(\dfrac{n}{m}\right)^{n-1}\)
\(u_n = m^{2}n^{3} \cdot \dfrac{n^{n-1}}{m^{n-1}}\)
\(u_n = m^{2-(n-1)}n^{3+(n-1)}\)
\(u_n = m^{3-n}n^{n+2}\)
Answer: the nth term \(u_n = m^{3-n}n^{n+2}\)
Divide the second term by the first term.
\(r = \dfrac{q^{2}}{p^{2}q}\)
\(r = p^{0-2}q^{2-1}\)
\(r = \dfrac{q}{p^{2}}\)
Multiply the second term by r. The product is the third term.
\(q^{2} \times \left(\dfrac{q}{p^{2}}\right) = p^{-2}q^{3}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = p^{2}q \times \left(\dfrac{q}{p^{2}}\right)^{n-1}\)
\(u_n = p^{2}q \cdot \dfrac{q^{n-1}}{p^{2(n-1)}}\)
\(u_n = p^{2-2(n-1)}q^{1+(n-1)}\)
\(u_n = p^{4-2n}q^{n}\)
Answer: the nth term \(u_n = p^{4-2n}q^{n}\)
Divide the second term by the first term.
\(r = \dfrac{6y^{2}}{2xy}\)
\(r = \dfrac{6}{2}x^{0-1}y^{2-1}\)
\(r = \dfrac{3y}{x}\)
Multiply the second term by r. The product is the third term.
\(6y^{2} \times \left(\dfrac{3y}{x}\right) = 18x^{-1}y^{3}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = 2xy \times \left(\dfrac{3y}{x}\right)^{n-1}\)
\(u_n = 2xy \cdot \dfrac{3^{n-1}y^{n-1}}{x^{n-1}}\)
\(u_n = 2 \cdot 3^{n-1}x^{1-(n-1)}y^{1+(n-1)}\)
\(u_n = 2 \cdot 3^{n-1}x^{2-n}y^{n}\)
Answer: the nth term \(u_n = 2 \cdot 3^{n-1}x^{2-n}y^{n}\)
Divide the second term by the first term.
\(r = \dfrac{xy^{3}}{x^{3}y^{2}}\)
\(r = x^{1-3}y^{3-2}\)
\(r = \dfrac{y}{x^{2}}\)
Multiply the second term by r. The product is the third term.
\(xy^{3} \times \left(\dfrac{y}{x^{2}}\right) = x^{-1}y^{4}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = x^{3}y^{2} \times \left(\dfrac{y}{x^{2}}\right)^{n-1}\)
\(u_n = x^{3}y^{2} \cdot \dfrac{y^{n-1}}{x^{2(n-1)}}\)
\(u_n = x^{3-2(n-1)}y^{2+(n-1)}\)
\(u_n = x^{5-2n}y^{n+1}\)
Answer: the nth term \(u_n = x^{5-2n}y^{n+1}\)
Simplify uₙ, then substitute the term number.
Divide the second term by the first term.
\(r = \dfrac{x^{-1}}{x^{-2}y}\)
\(r = x^{(-1)-(-2)}y^{0-1}\)
\(r = \dfrac{x}{y}\)
Multiply the second term by r. The product is the third term.
\(x^{-1} \times \left(\dfrac{x}{y}\right) = y^{-1}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = x^{-2}y \times \left(\dfrac{x}{y}\right)^{n-1}\)
\(u_n = x^{-2}y \cdot \dfrac{x^{n-1}}{y^{n-1}}\)
\(u_n = x^{-2+(n-1)}y^{1-(n-1)}\)
\(u_n = x^{n-3}y^{2-n}\)
Substitute \(n = 5\).
\(u_{5} = x^{5-3}y^{2-5}\)
\(u_{5} = x^{2}y^{-3}\)
\(u_{5} = \dfrac{x^{2}}{y^{3}}\)
Answer: the nth term \(u_n = x^{n-3}y^{2-n}\) and the 5th term = \(\dfrac{x^{2}}{y^{3}}\)
Divide the second term by the first term.
\(r = \dfrac{abc}{a^{2}b}\)
\(r = a^{1-2}b^{1-1}c^{1-0}\)
\(r = \dfrac{c}{a}\)
Multiply the second term by r. The product is the third term.
\(abc \times \left(\dfrac{c}{a}\right) = bc^{2}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = a^{2}b \times \left(\dfrac{c}{a}\right)^{n-1}\)
\(u_n = a^{2}b \cdot \dfrac{c^{n-1}}{a^{n-1}}\)
\(u_n = a^{2-(n-1)}bc^{n-1}\)
\(u_n = a^{3-n}bc^{n-1}\)
Substitute \(n = 4\).
\(u_{4} = a^{3-4}bc^{4-1}\)
\(u_{4} = a^{-1}bc^{3}\)
\(u_{4} = \dfrac{bc^{3}}{a}\)
Answer: the nth term \(u_n = a^{3-n}bc^{n-1}\) and the 4th term = \(\dfrac{bc^{3}}{a}\)
Divide the second term by the first term.
\(r = \dfrac{6x}{4x^{2}}\)
\(r = \dfrac{6}{4}x^{1-2}\)
\(r = \dfrac{3}{2x}\)
Multiply the second term by r. The product is the third term.
\(6x \times \left(\dfrac{3}{2x}\right) = 9\)
\(u_n = u_1 r^{n-1}\)
\(u_n = 4x^{2} \times \left(\dfrac{3}{2x}\right)^{n-1}\)
\(u_n = 4x^{2} \cdot \dfrac{3^{n-1}}{2^{n-1}x^{n-1}}\)
\(u_n = 4 \cdot \left(\dfrac{3}{2}\right)^{n-1}x^{2-(n-1)}\)
\(u_n = 2^{3-n} \cdot 3^{n-1}x^{3-n}\)
Substitute \(n = 4\).
\(u_{4} = 2^{3-4} \cdot 3^{4-1}x^{3-4}\)
\(u_{4} = \dfrac{27}{2}x^{-1}\)
\(u_{4} = \dfrac{27}{2x}\)
Answer: the nth term \(u_n = 2^{3-n} \cdot 3^{n-1}x^{3-n}\) and the 4th term = \(\dfrac{27}{2x}\)
Divide the second term by the first term.
\(r = \dfrac{6x^{2}}{-2x}\)
\(r = -\dfrac{6}{2}x^{2-1}\)
\(r = -3x\)
Multiply the second term by r. The product is the third term.
\(6x^{2} \times \left(-3x\right) = -18x^{3}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = -2x \times \left(-3x\right)^{n-1}\)
\(u_n = -2x \cdot \left(-3\right)^{n-1}x^{n-1}\)
\(u_n = -2 \cdot \left(-3\right)^{n-1}x^{1+(n-1)}\)
\(u_n = -2 \cdot \left(-3\right)^{n-1}x^{n}\)
Substitute \(n = 5\).
\(u_{5} = -2 \cdot \left(-3\right)^{5-1}x^{5}\)
\(u_{5} = -162x^{5}\)
Answer: the nth term \(u_n = -2 \cdot \left(-3\right)^{n-1}x^{n}\) and the 5th term = \(-162x^{5}\)
Divide the second term by the first term.
\(r = \dfrac{y^{-2}}{xy^{-1}}\)
\(r = x^{0-1}y^{(-2)-(-1)}\)
\(r = \dfrac{1}{xy}\)
Multiply the second term by r. The product is the third term.
\(y^{-2} \times \left(\dfrac{1}{xy}\right) = x^{-1}y^{-3}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = xy^{-1} \times \left(\dfrac{1}{xy}\right)^{n-1}\)
\(u_n = xy^{-1} \cdot \dfrac{1}{x^{n-1}y^{n-1}}\)
\(u_n = x^{1-(n-1)}y^{-1-(n-1)}\)
\(u_n = x^{2-n}y^{-n}\)
Substitute \(n = 4\).
\(u_{4} = x^{2-4}y^{-4}\)
\(u_{4} = x^{-2}y^{-4}\)
\(u_{4} = \dfrac{1}{x^{2}y^{4}}\)
Answer: the nth term \(u_n = x^{2-n}y^{-n}\) and the 4th term = \(\dfrac{1}{x^{2}y^{4}}\)
Divide the second term by the first term.
\(r = \dfrac{b^{4}}{ab^{3}}\)
\(r = a^{0-1}b^{4-3}\)
\(r = \dfrac{b}{a}\)
Multiply the second term by r. The product is the third term.
\(b^{4} \times \left(\dfrac{b}{a}\right) = a^{-1}b^{5}\)
\(u_n = u_1 r^{n-1}\)
\(u_n = ab^{3} \times \left(\dfrac{b}{a}\right)^{n-1}\)
\(u_n = ab^{3} \cdot \dfrac{b^{n-1}}{a^{n-1}}\)
\(u_n = a^{1-(n-1)}b^{3+(n-1)}\)
\(u_n = a^{2-n}b^{n+2}\)
Substitute \(n = 6\).
\(u_{6} = a^{2-6}b^{6+2}\)
\(u_{6} = a^{-4}b^{8}\)
\(u_{6} = \dfrac{b^{8}}{a^{4}}\)
Answer: the nth term \(u_n = a^{2-n}b^{n+2}\) and the 6th term = \(\dfrac{b^{8}}{a^{4}}\)
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