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IB Mathematics SL 1.3 Geometric sequences and series AI SL Paper 2 - Exam Style Questions - New Syllabus

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Question

The admissions team at a new College are trying to predict the number of student applications they will receive each year.
Let \(n\) be the number of years that the College has been open. The admissions team collect the following data for the first two years.
Year, \(n\)Number of applications received in year \(n\)
112 300
212 669
(a) Calculate the percentage increase in applications from the first year to the second year. [2]
It is assumed that the number of students that apply to the College each year will follow a geometric sequence, \(u_n\).
(b) (i) Write down the common ratio of the sequence.
     (ii) Find an expression for \(u_n\).
     (iii) Find the number of student applications the College expects to receive when \(n=11\). Express your answer to the nearest integer. [4]
In the first year there were 10 380 places at the College available for applicants. The admissions team announce that the number of places available will increase by 600 every year.
Let \(v_n\) represent the number of places available at the College in year \(n\).
(c) Write down an expression for \(v_n\). [2]
For the first 10 years that the College is open, all places are filled. Students who receive a place each pay an \$80 acceptance fee.
(d) Calculate the total amount of acceptance fees paid to the College in the first 10 years. [3]
When \(n=k\), the number of places available will, for the first time, exceed the number of students applying.
(e) Find \(k\). [3]
(f) State whether, for all \(n>k\), the College will have places available for all applicants. Justify your answer. [2]
▶️ Answer / Explanation (Detailed working)
Markscheme-style solution

(a) Percentage increase from year 1 to year 2:

\[ \frac{12\,669-12\,300}{12\,300}\times 100\% \;=\;\frac{369}{12\,300}\times 100\% \;=\; \boxed{3\%}. \]

(b)

(i) \(\displaystyle r=\frac{12\,669}{12\,300}=\boxed{1.03}\).
(ii) \(\displaystyle \boxed{u_n=12\,300\,(1.03)^{\,n-1}}\).
(iii) \(\displaystyle u_{11}=12\,300\,(1.03)^{10}\approx 12\,300\times 1.343916379\approx 16\,530.171\). Nearest integer: \(\boxed{16\,530}\).

(c) Arithmetic sequence for places with \(v_1=10\,380\) and common difference \(d=600\):

\[ \boxed{v_n=10\,380+600(n-1)}\quad\text{(equivalently, }v_n=600n+9\,780\text{)}. \]

(d) For the first 10 years all places are filled, so paying students each year \(=v_n\). Sum of first 10 terms of an arithmetic sequence:

\[ S_{10}=\frac{10}{2}\big(2\cdot 10\,380+9\cdot 600\big) =5\,(20\,760+5\,400) =5\times 26\,160=130\,800. \] Total acceptance fees \(= \$80\times 130\,800=\boxed{\$10\,464\,000}.\)

(e) First year when places exceed applications: smallest integer \(k\) with \(v_k>u_k\).

\(v_{12}=10\,380+600\cdot 11=16\,980,\quad u_{12}=12\,300(1.03)^{11}\approx 17\,026.1\Rightarrow v_{12}<u_{12}\).
\(v_{13}=10\,380+600\cdot 12=17\,580,\quad u_{13}=12\,300(1.03)^{12}\approx 17\,536.9\Rightarrow v_{13}>u_{13}\).
Hence \(\boxed{k=13}\).

(f) No. \(u_n\) grows geometrically (ratio \(1.03>1\)) while \(v_n\) grows linearly. Eventually \(u_n\) will exceed \(v_n\) again. For example,

\[ u_{24}=12\,300(1.03)^{23}\approx 24\,275.1,\qquad v_{24}=10\,380+600\cdot 23=24\,180, \] so \(u_{24}>v_{24}\). Therefore it is not true that for all \(n>k\) there will be places for all applicants.
Total Marks: 16
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