Geometric Sequences I
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Consider the sequence: \(3, 6, 12, 24, \ldots\)
(a)Find the \(n\)th term.
(b)Find the 10th term.
(c)Is 768 in the sequence?
(a)
Answer
\(u_n=3 imes2^{n-1}\)
(c)
Solve \(3 imes2^{n-1}=768\)
\(2^{n-1}=256=2^8\Rightarrow n=9\) ✓
Question 2
A geometric sequence has \(u_3=45\) and \(u_6=1215\). Find the first term and common ratio.
Divide
\(\dfrac{u_6}{u_3}=r^3=27\Rightarrow r=3\)
Back-substitute
\(u_1\cdot9=45\Rightarrow u_1=5\)
Question 3
Find \(k\), given that the following are geometric sequences:
(a)\(2, k, 18\)
(b)\(k, 3k+4, 9k+4\)
(a)
Middle term squared = product of outer terms
\(k^2=36\Rightarrow k=\pm6\)
Answer
\(k=6\) or \(k=-6\)
(b)
Common ratio condition
\((3k+4)^2=k(9k+4)\Rightarrow 9k^2+24k+16=9k^2+4k\Rightarrow 20k=-16\)
Question 4
A geometric sequence has first term 5 and \(u_n=5 imes3^{n-1}\). Which is the first term to exceed 1000?
Solve
\(5 imes3^{n-1}>1000\Rightarrow 3^{n-1}>200\Rightarrow (n-1)\log3>\log200\Rightarrow n>1+ frac{\log200}{\log3}pprox5.86\)
Answer
\(n=6\), \(u_6=1215\)
Question 5
The second and fifth terms of a geometric sequence are 12 and \( frac{3}{2}\). Find all possible values of the common ratio and the first term.
Divide
\(r^3=\dfrac{u_5}{u_2}=\dfrac{3/2}{12}= frac{1}{8}\Rightarrow r= frac{1}{2}\)
First term
\(u_1\cdot r=12\Rightarrow u_1=24\)
Answer
\(r= frac{1}{2},\;u_1=24\)
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