Sequences Mixed — Past Paper Style (HL)
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Find \(k\) such that \(\displaystyle\sum_{r=1}^{k}(2r+1)=120\).
Expand sum
\(\displaystyle\sum_{r=1}^k(2r+1)=2\cdot\dfrac{k(k+1)}{2}+k=k^2+2k=120\Rightarrow k^2+2k-120=0\Rightarrow(k+12)(k-10)=0\)
Question 2
The 3rd term of a geometric sequence is 4 and the 7th term is \( frac{1}{4}\). Find the sum to infinity.
Find \(r\)
\(r^4=\dfrac{1/4}{4}= frac{1}{16}\Rightarrow r=\pm frac{1}{2}\)
Find \(u_1\)
With \(r= frac{1}{2}\): \(u_1\cdot frac{1}{4}=4\Rightarrow u_1=16\). With \(r=- frac{1}{2}\): same \(u_1=16\)
Sum to infinity
\(S_\infty=\dfrac{16}{1-(\pm frac{1}{2})}\)
Answer
\(S_\infty=32\) (if \(r= frac{1}{2}\)) or \(\dfrac{32}{3}\) (if \(r=- frac{1}{2}\))
Question 3
Prove by induction that \(\displaystyle\sum_{r=1}^n r = \dfrac{n(n+1)}{2}\).
Base case
\(n=1\): LHS\(=1\); RHS\(= frac{1\cdot2}{2}=1\) ✓
Inductive step
Assume true for \(n=k\). For \(n=k+1\): \(\displaystyle\sum_{r=1}^{k+1}r=\dfrac{k(k+1)}{2}+(k+1)=\dfrac{k(k+1)+2(k+1)}{2}=\dfrac{(k+1)(k+2)}{2}\) ✓
Conclusion
True by PMI for all \(n\in\mathbb{Z}^+\). ■
Question 4
An arithmetic and a geometric sequence both start \(1, a, b, \ldots\) Find \(a\) and \(b\) for the geometric sequence such that the sum to infinity exists, given the arithmetic common difference is 3.
Arithmetic
\(a=1+3=4,\;b=7\)
Geometric with same \(a,b\) not possible if it must also start 1
Geometric: \(r=a,\;b=a^2\). For \(|r|<1\): need \(|a|<1\). With arithmetic \(d=3\): \(a=4\) — not valid for convergence. Take general geometric starting 1: \(r=a\), need \(|a|<1\).
Answer
The sequences cannot share the same \(a=4\) and converge; if geometric has \(r=a\) with \(|a|<1\), any valid \(a\in(-1,1)\setminus\{0\}\) works. Contradiction — arithmetic gives \(a=4\), geometric with \(|r|<1\) requires \(|a|<1\).
Question 5
The sum of an infinite geometric series is twice its first term. Find the common ratio. Hence find the first term if the sum of the first 5 terms is 31.
Condition
\(\dfrac{u_1}{1-r}=2u_1\Rightarrow1-r= frac{1}{2}\Rightarrow r= frac{1}{2}\)
Sum of 5 terms
\(u_1\cdot\dfrac{1-(1/2)^5}{1/2}=2u_1\cdot\dfrac{31}{32}=31\Rightarrow u_1=16\)
Answer
\(r= frac{1}{2},\;u_1=16\)
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