IBDP Maths AA HL – SL 1.2 Arithmetic Sequences & Series HL Paper 1 Exam Style Questions
IB MATHEMATICS AA HL – Practice Questions- Paper1-All Topics
Topic : SL 1.2 Arithmetic Sequences & Series HL Paper 1
Topic 1- Number and algebra- Weightage : 15 %
All Questions for Topic : SL 1.2 Arithmetic Sequences & Series HL Paper 1. Use of the formulae for the nth term and the sum of the first n terms of the sequence. Use of sigma notation for sums of arithmetic sequences. Applications. Analysis, interpretation and prediction where a model is not perfectly arithmetic in real life.. approximate common differences.
[No-Calc][Easy] [Maximum Mark:7]
Question: SL 1.2 Arithmetic Sequences & Series HL Paper 1
Consider the series ln x p + + ln x + \(\frac{1}{3}In x + ….,\) where x ∈ R, x > 1 and p ∈ R, p ≠ 0.
(a) Consider the case where the series is geometric.
(i) Show that p = ± \(\frac{1}{\sqrt{3}}\)
▶️Answer/Explanation
Ans: EITHER
attempt to use a ratio from consecutive terms
\(\frac{p In x}{In x} = \frac{\frac{1}{3}In x}{p In x} OR \frac{1}{3}In x = (In x)r^{2} OR p In x = In x \left ( \frac{1}{3p} \right )\)
Note: Candidates may use \(In x^{1} + In x^{p} + Inx^{\frac{1}{3}} + …..\) and consider the powers of x in geometric sequence.
Award M1 for \(\frac{p}{1} = \frac{\frac{1}{3}}{p}.\)
OR
r = p and r2 = \(\frac{1}{3}\)
THEN
\(p^{2}= \frac{1}{3} OR r = \pm \frac{1}{\sqrt{3}}\)
\(p = \pm \frac{1}{\sqrt{3}}\)
Note: Award M0A0 for \(r^{2} = \frac{1}{3} or p^{2} = \frac{1}{3}\) with no other working seen.
(ii) Hence or otherwise, show that the series is convergent.
▶️Answer/Explanation
Ans: EITHER
\(since, \left | p \right |= \frac{1}{\sqrt{3}} and \frac{1}{\sqrt{3}}<1\)
OR
\(since, \left | p \right |= \frac{1}{\sqrt{3}} and -1<p<1\)
THEN
⇒the geometric series converges.
Note: Accept r instead of p .
Award RO if both values of p not considered.
(iii) Given that p > 0 and S∞ = 3 + \(\sqrt{3}\) , find the value of x .
▶️Answer/Explanation
Ans: \(\frac{In x}{1-\frac{1}{\sqrt{3}}} (=3 + \sqrt{3})\)
\(In x = 3 – \frac{3}{\sqrt{3}}+ \sqrt{3} – \frac{\sqrt{3}}{\sqrt{3}}\) OR \(In x = 3 – \sqrt{3} + \sqrt{3} – 1 (\Rightarrow In x =2)\)
x = e2
(b) Now consider the case where the series is arithmetic with common difference d.
(i) Show that p = \(\frac{2}{3}\)
▶️Answer/Explanation
Ans: METHOD 1
attempt to find a difference from consecutive terms or from u2 correct equation
\(p In x – In x = \frac{1}{3}In x – p In x OR \frac{1}{3}In x = In x +2 (p In x – In x)\)
Note: Candidates may use \(In x^{1} + In x^{p} + In x^{\frac{1}{3}} + ……\) and consider the powers of x in arithmetic sequence.
Award M1A1 for \(p – 1 = \frac{1}{3} – p.\)
\(2p In x = \frac{4}{3}In x \left ( \Rightarrow 2p = \frac{4}{3} \right )\)
\(p = \frac{2}{3}\)
METHOD 2
attempt to use arithmetic mean \(u_{2} = \frac{u_{1}+u_{3}}{2}\)
\(p In x = \frac{In x + \frac{1}{3}In x}{2}\)
\(2p In x = \frac{4}{3}In x \left ( \Rightarrow 2p = \frac{4}{3} \right )\)
\(p = \frac{2}{3}\)
(ii) Write down d in the form k ln x , where k ∈ R.
▶️Answer/Explanation
Ans: \(d = -\frac{1}{3}In x\)
(iii) The sum of the first n terms of the series is ln \(\left ( \frac{1}{x^{3}} \right )\)
Find the value of n.
▶️Answer/Explanation
Ans: METHOD 1
\(S_{n} = \frac{n}{2}\left [ 2 In x +(n-1)\times \left ( -\frac{1}{3}In x \right ) \right ]\)
attempt to substitute into Sn and equate to \(In \left ( \frac{1}{x^{3}} \right )\)
\(\frac{n}{2}\left \lfloor 2In x + (n-1)\times \left ( -\frac{1}{3} In x \right ) \right \rfloor = In \left ( \frac{1}{x^{3}} \right )\)
\(In\left ( \frac{1}{x^{3}} \right ) = -In x^{3}\left ( = In x^{-3} \right )\)
= −3ln x
correct working with Sn (seen anywhere)
\(\frac{n}{2}\left \lfloor 2 In x – \frac{n}{3} In x + \frac{1}{3}In x\right \rfloor OR nIn x – \frac{n(n-1)}{6}In x OR \frac{n}{2}\left ( In x + \left ( \frac{4-n}{3} \right )In x\right )\)
correct equation without ln x
\(\frac{n}{2}\left ( \frac{7}{3} – \frac{n}{3} \right ) = -3 OR n-\frac{n(n-1)}{6} = -3\) (or equivalent)
Note: Award as above if the series \(1 + p + \frac{1}{3}+ ….\) is considered leading to \(\frac{n}{2}\left ( \frac{7}{3} -\frac{n}{3}\right ) = -3.\)
attempt to form a quadratic = 0
n2 -7n − 18 = 0
attempt to solve their quadratic
(n – 9) (n + 2) = 0
n = 9
Question:
Consider the series ln x p + + ln x + \(\frac{1}{3}In x + ….,\) where x ∈ R, x > 1 and p ∈ R, p ≠ 0.
(a) Consider the case where the series is geometric.
(i) Show that p = ± \(\frac{1}{\sqrt{3}}\)
▶️Answer/Explanation
Ans: EITHER
attempt to use a ratio from consecutive terms
\(\frac{p In x}{In x} = \frac{\frac{1}{3}In x}{p In x} OR \frac{1}{3}In x = (In x)r^{2} OR p In x = In x \left ( \frac{1}{3p} \right )\)
Note: Candidates may use \(In x^{1} + In x^{p} + Inx^{\frac{1}{3}} + …..\) and consider the powers of x in geometric sequence.
Award M1 for \(\frac{p}{1} = \frac{\frac{1}{3}}{p}.\)
OR
r = p and r2 = \(\frac{1}{3}\)
THEN
\(p^{2}= \frac{1}{3} OR r = \pm \frac{1}{\sqrt{3}}\)
\(p = \pm \frac{1}{\sqrt{3}}\)
Note: Award M0A0 for \(r^{2} = \frac{1}{3} or p^{2} = \frac{1}{3}\) with no other working seen.
(ii) Hence or otherwise, show that the series is convergent.
▶️Answer/Explanation
Ans: EITHER
\(since, \left | p \right |= \frac{1}{\sqrt{3}} and \frac{1}{\sqrt{3}}<1\)
OR
\(since, \left | p \right |= \frac{1}{\sqrt{3}} and -1<p<1\)
THEN
⇒the geometric series converges.
Note: Accept r instead of p .
Award RO if both values of p not considered.
(iii) Given that p > 0 and S∞ = 3 + \(\sqrt{3}\) , find the value of x .
▶️Answer/Explanation
Ans: \(\frac{In x}{1-\frac{1}{\sqrt{3}}} (=3 + \sqrt{3})\)
\(In x = 3 – \frac{3}{\sqrt{3}}+ \sqrt{3} – \frac{\sqrt{3}}{\sqrt{3}}\) OR \(In x = 3 – \sqrt{3} + \sqrt{3} – 1 (\Rightarrow In x =2)\)
x = e2
(b) Now consider the case where the series is arithmetic with common difference d.
(i) Show that p = \(\frac{2}{3}\)
▶️Answer/Explanation
Ans: METHOD 1
attempt to find a difference from consecutive terms or from u2 correct equation
\(p In x – In x = \frac{1}{3}In x – p In x OR \frac{1}{3}In x = In x +2 (p In x – In x)\)
Note: Candidates may use \(In x^{1} + In x^{p} + In x^{\frac{1}{3}} + ……\) and consider the powers of x in arithmetic sequence.
Award M1A1 for \(p – 1 = \frac{1}{3} – p.\)
\(2p In x = \frac{4}{3}In x \left ( \Rightarrow 2p = \frac{4}{3} \right )\)
\(p = \frac{2}{3}\)
METHOD 2
attempt to use arithmetic mean \(u_{2} = \frac{u_{1}+u_{3}}{2}\)
\(p In x = \frac{In x + \frac{1}{3}In x}{2}\)
\(2p In x = \frac{4}{3}In x \left ( \Rightarrow 2p = \frac{4}{3} \right )\)
\(p = \frac{2}{3}\)
(ii) Write down d in the form k ln x , where k ∈ R.
▶️Answer/Explanation
Ans: \(d = -\frac{1}{3}In x\)
(iii) The sum of the first n terms of the series is ln \(\left ( \frac{1}{x^{3}} \right )\) Find the value of n.
▶️Answer/Explanation
Ans: METHOD 1
\(S_{n} = \frac{n}{2}\left [ 2 In x +(n-1)\times \left ( -\frac{1}{3}In x \right ) \right ]\)
attempt to substitute into Sn and equate to \(In \left ( \frac{1}{x^{3}} \right )\)
\(\frac{n}{2}\left \lfloor 2In x + (n-1)\times \left ( -\frac{1}{3} In x \right ) \right \rfloor = In \left ( \frac{1}{x^{3}} \right )\)
\(In\left ( \frac{1}{x^{3}} \right ) = -In x^{3}\left ( = In x^{-3} \right )\)
= −3ln x
correct working with Sn (seen anywhere)
\(\frac{n}{2}\left \lfloor 2 In x – \frac{n}{3} In x + \frac{1}{3}In x\right \rfloor OR nIn x – \frac{n(n-1)}{6}In x OR \frac{n}{2}\left ( In x + \left ( \frac{4-n}{3} \right )In x\right )\)
correct equation without ln x
\(\frac{n}{2}\left ( \frac{7}{3} – \frac{n}{3} \right ) = -3 OR n-\frac{n(n-1)}{6} = -3\) (or equivalent)
Note: Award as above if the series \(1 + p + \frac{1}{3}+ ….\) is considered leading to \(\frac{n}{2}\left ( \frac{7}{3} -\frac{n}{3}\right ) = -3.\)
attempt to form a quadratic = 0
n2 -7n − 18 = 0
attempt to solve their quadratic
(n – 9) (n + 2) = 0
n = 9
Question:
The nth term of an arithmetic sequence is given by un = 15 – 3n.
(a) State the value of the first term, u1 .
▶️Answer/Explanation
Ans: u1 =12
(b) Given that the nth term of this sequence is -33, find the value of n.
▶️Answer/Explanation
Ans: 15 – 3n = -33
n = 16
(c) Find the common difference, d.
▶️Answer/Explanation
Ans: valid approach to find d
u2 – u1 = 9 – 12 OR recognize gradient is −3 OR attempts to solve
-33 = 12 + 15 d
d = -3
Question:
Calculate the sums
(a) 7 +10 +13 +16 +……….+157
▶️Answer/Explanation
Ans: 4182
(b) \(\sum_{r=1}^{51}(3r+4)\)
▶️Answer/Explanation
Ans: 4182
Question: SL 1.2 Arithmetic Sequences & Series HL Paper 1
Consider an arithmetic sequence where u8 = S8 = 8 . Find the value of the first term, u1 , and the value of the common difference, d .
▶️Answer/Explanation
Ans: 
Question:
Find the value of k if \({\sum\limits_{r = 1}^\infty{k\left( {\frac{1}{3}} \right)}^r} = 7\).
▶️Answer/Explanation
Ans: \({u_1} = \frac{1}{3}k{\text{ , }}r = \frac{1}{3}\) (A1) (A1)
\(7 = \frac{{\frac{1}{3}k}}{{1 – \frac{1}{3}}}\) M1
\(k = 14\) A1
[4 marks]
Question:
a. Let \(\{ {u_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be an arithmetic sequence with first term equal to \(a\) and common difference of \(d\), where \(d \ne 0\). Let another sequence \(\{ {v_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be defined by \({v_n} = {2^{{u_n}}}\).
(i) Show that \(\frac{{{v_{n + 1}}}}{{{v_n}}}\) is a constant.
▶️Answer/Explanation
Ans: METHOD 1
\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{{u_{n + 1}}}}}}{{{2^{{u_n}}}}}\) M1
\( = {2^{{u_{n + 1}} – {u_n}}} = {2^d}\) A1
METHOD 2
\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{a + nd}}}}{{{2^{a + (n – 1)d}}}}\) M1
\( = {2^d}\) A1
(ii) Write down the first term of the sequence \(\{ {v_n}\} \).
▶️Answer/Explanation
Ans: \( = {2^a}\) A1
Note: Accept \( = {2^{{u_1}}}\).
(iii) Write down a formula for \({v_n}\) in terms of \(a\), \(d\) and \(n\).[4]
▶️Answer/Explanation
Ans: EITHER
\({v_n}\) is a GP with first term \({2^a}\) and common ratio \({2^d}\)
\({v_n} = {2^a}{({2^d})^{(n – 1)}}\)
OR
\({u_n} = a + (n – 1)d\) as it is an AP
THEN
\({v_n} = {2^a}^{ + (n – 1)d}\) A1
[4 marks]
b. Let \({S_n}\) be the sum of the first \(n\) terms of the sequence \(\{ {v_n}\} \).
(i) Find \({S_n}\), in terms of \(a\), \(d\) and \(n\).
▶️Answer/Explanation
Ans: \({S_n} = \frac{{{2^a}\left( {{{({2^d})}^n} – 1} \right)}}{{{2^d} – 1}} = \frac{{{2^a}({2^{dn}} – 1)}}{{{2^d} – 1}}\) M1A1
Note: Accept either expression.
(ii) Find the values of \(d\) for which \(\sum\limits_{i = 1}^\infty {{v_i}} \) exists.
You are now told that \(\sum\limits_{i = 1}^\infty {{v_i}} \) does exist and is denoted by \({S_\infty }\).
▶️Answer/Explanation
Ans: for sum to infinity to exist need \( – 1 < {2^d} < 1\) R1
\( \Rightarrow \log {2^d} < 0 \Rightarrow d\log 2 < 0 \Rightarrow d < 0\) (M1)A1
Note: Also allow graph of \({2^d}\).
(iii) Write down \({S_\infty }\) in terms of \(a\) and \(d\) .
▶️Answer/Explanation
Ans: \({S_\infty } = \frac{{{2^a}}}{{1 – {2^d}}}\) A1
(iv) Given that \({S_\infty } = {2^{a + 1}}\) find the value of \(d\) .[8]
▶️Answer/Explanation
Ans: \(\frac{{{2^a}}}{{1 – {2^d}}} = {2^{a + 1}} \Rightarrow \frac{1}{{1 – {2^d}}} = 2\) M1
\( \Rightarrow 1 = 2 – {2^{d + 1}} \Rightarrow {2^{d + 1}} = 1\)
\( \Rightarrow d = – 1\) A1
[8 marks]
c. Let \(\{ {w_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be a geometric sequence with first term equal to \(p\) and common ratio \(q\), where \(p\) and \(q\) are both greater than zero. Let another sequence \(\{ {z_n}\} \) be defined by \({z_n} = \ln {w_n}\).
Find \(\sum\limits_{i = 1}^n {{z_i}} \) giving your answer in the form \(\ln k\) with \(k\) in terms of \(n\), \(p\) and \(q\).[6]
▶️Answer/Explanation
Ans: METHOD 1
\({w_n} = p{q^{n – 1}},{\text{ }}{z_n} = \ln p{q^{n – 1}}\) (A1)
\({z_n} = \ln p + (n – 1)\ln q\) M1A1
\({z_{n + 1}} – {z_n} = (\ln p + n\ln q) – (\ln p + (n – 1)\ln q) = \ln q\)
which is a constant so this is an AP
(with first term \(\ln p\) and common difference \(\ln q\))
\(\sum\limits_{i = 1}^n {{z_i} = \frac{n}{2}\left( {2\ln p + (n – 1)\ln q} \right)} \) M1
\( = n\left( {\ln p + \ln {q^{\left( {\frac{{n – 1}}{2}} \right)}}} \right) = n\ln \left( {p{q^{\left( {\frac{{n – 1}}{2}} \right)}}} \right)\) (M1)
\( = \ln \left( {{p^n}{q^{\frac{{n(n – 1)}}{2}}}} \right)\) A1
METHOD 2
\(\sum\limits_{i = 1}^n {{z_i} = \ln p + \ln pq + \ln p{q^2} + \ldots + \ln p{q^{n – 1}}} \) (M1)A1
\( = \ln \left( {{p^n}{q^{\left( {1 + 2 + 3 + \ldots + (n – 1)} \right)}}} \right)\) (M1)A1
\( = \ln \left( {{p^n}{q^{\frac{{n(n – 1)}}{2}}}} \right)\) (M1)A1
[6 marks]
Total [18 marks]
Examiners report
a. Method of first part was fine but then some algebra mistakes often happened. The next two parts were generally good.
b. Given that (a) indicated that there was a common ratio a disappointing number thought it was an AP. Although some good answers in the next parts, there was also some poor notational misunderstanding with the sum to infinity still involving \(n\).
c. Not enough candidates realised that this was an AP.
Question:
a. Find the sum of the infinite geometric sequence 27, −9, 3, −1, … .[3]
▶️Answer/Explanation
Ans: \(r = – \frac{1}{3}\) (A1)
\({S_\infty } = \frac{{27}}{{1 + \frac{1}{3}}}\) M1
\({S_\infty } = \frac{{81}}{4}\,\,\,\,\,( = 20.25)\) A1 N1
[3 marks]
b. Use mathematical induction to prove that for n ∈ Z+, \(a+ar+ar^{2}+….+ar^{n-1}=\frac{a(1-r^{n})}{1-r}.\) [7]
▶️Answer/Explanation
Ans: Attempting to show that the result is true for n = 1 M1
LHS = a and \({\text{RHS}} = \frac{{a(1 – r)}}{{1 – r}} = a\) A1
Hence the result is true for n = 1
Assume it is true for n = k
\(a + ar + a{r^2} + … + a{r^{k – 1}} = \frac{{a(1 – rk)}}{{1 – r}}\) M1
Consider n = k + 1:
\(a + ar + a{r^2} + … + a{r^{k – 1}} + a{r^k} = \frac{{a(1 – {r^k})}}{{1 – r}} + a{r^k}\) M1
\( = \frac{{a(1 – {r^k}) + a{r^k}(1 – r)}}{{1 – r}}\)
\( = \frac{{a – a{r^k} + a{r^k} – a{r^{k + 1}}}}{{1 – r}}\) A1
Note: Award A1 for an equivalent correct intermediate step.
\( = \frac{{a – a{r^{k + 1}}}}{{1 – r}}\)
\( = \frac{{a(1 – {r^{k + 1}})}}{{1 – r}}\) A1
Note: Illogical attempted proofs that use the result to be proved would gain M1A0A0 for the last three above marks.
The result is true for \(n = k \Rightarrow \) it is true for \(n = k + 1\) and as it is true for \(n = 1\), the result is proved by mathematical induction. R1 N0
Note: To obtain the final R1 mark a reasonable attempt must have been made to prove the k + 1 step.
[7 marks]