CAIE P1 2024 June — Question 8 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeMixed arithmetic and geometric
DifficultyModerate -0.3 This is a straightforward two-part question on arithmetic and geometric progressions requiring standard techniques. Part (a) uses the constant difference property to find p, then applies the nth term formula. Part (b) uses the constant ratio property to find q (solving a quadratic), then applies the sum to infinity formula. While it requires careful algebra and multiple steps for the marks available, it involves only routine procedures with no novel insight or problem-solving beyond textbook exercises, making it slightly easier than average.
Spec1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1
  1. The first three terms of an arithmetic progression are \(25\), \(4p - 1\) and \(13 - p\), where \(p\) is a constant. Find the value of the tenth term of the progression. [4]
  2. The first three terms of a geometric progression are \(25\), \(4q - 1\) and \(13 - q\), where \(q\) is a positive constant. Find the sum to infinity of the progression. [4]
Question 8:
AnswerMarks Guidance
8(a)24p12513p *M1
40
p
AnswerMarks
9A1
 40   74
d 4 125 
 
AnswerMarks Guidance
 9   9 DM1 Using their p to find d.
10th term = 259d49A1
Alternative Method for first 3 marks of Question 8(a)
AnswerMarks Guidance
d  4p26, d 145p , p + 2d = -12 Any two(*M1) Allow unsimplified or equivalent.
Solving simultaneously to find p or d(DM1)
 40 74
p  , d 
AnswerMarks
  9   9(A1)
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(b)4q12 2513q 16q217q3240
⇒M1
q416q810 q4
AnswerMarks Guidance
leading toM1 Solve 3 term quadratic with real solutions.
3
r
AnswerMarks Guidance
5A1 17
Ignore .
20
25 125
Sum to infinity = 
3 2
1
AnswerMarks Guidance
5A1 Ignore extra solution.
SC B1 if no method shown for solving quadratic.
Alternative Method for Question 8(b)
AnswerMarks Guidance
25r = 4q – 1, 25r2 = 13 – q leading to 100r2 + 25r – 51 = 0(M1)
5r320r170(M1) Solve 3 term quadratic with real solutions.
3
r =
AnswerMarks Guidance
5(A1) 17
Ignore .
20
25 125
Sum to infinity = 
3 2
1
AnswerMarks Guidance
5(A1) Ignore extra solution.
SC B1 if no method shown for solving quadratic.
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 8:
--- 8(a) ---
8(a) | 24p12513p | *M1
40
p
9 | A1
 40   74
d 4 125 
 
 9   9  | DM1 | Using their p to find d.
10th term = 259d49 | A1
Alternative Method for first 3 marks of Question 8(a)
d  4p26, d 145p , p + 2d = -12 Any two | (*M1) | Allow unsimplified or equivalent.
Solving simultaneously to find p or d | (DM1)
 40 74
p  , d 
  9   9 | (A1)
4
Question | Answer | Marks | Guidance
--- 8(b) ---
8(b) | 4q12 2513q 16q217q3240
⇒ | M1
q416q810 q4
leading to | M1 | Solve 3 term quadratic with real solutions.
3
r
5 | A1 | 17
Ignore .
20
25 125
Sum to infinity = 
3 2
1
5 | A1 | Ignore extra solution.
SC B1 if no method shown for solving quadratic.
Alternative Method for Question 8(b)
25r = 4q – 1, 25r2 = 13 – q leading to 100r2 + 25r – 51 = 0 | (M1)
5r320r170 | (M1) | Solve 3 term quadratic with real solutions.
3
r =
5 | (A1) | 17
Ignore .
20
25 125
Sum to infinity = 
3 2
1
5 | (A1) | Ignore extra solution.
SC B1 if no method shown for solving quadratic.
4
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item The first three terms of an arithmetic progression are $25$, $4p - 1$ and $13 - p$, where $p$ is a constant.

Find the value of the tenth term of the progression. [4]

\item The first three terms of a geometric progression are $25$, $4q - 1$ and $13 - q$, where $q$ is a positive constant.

Find the sum to infinity of the progression. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q8 [8]}}
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