CAIE P1 2024 November — Question 6 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeSum to infinity from S_n ratio
DifficultyStandard +0.3 This is a standard geometric progression problem requiring manipulation of the sum formula and solving a quadratic equation. While it involves multiple steps (setting up S₄ and S₈, using the given ratio, solving for r, then finding S∞), the techniques are routine for A-level and the algebraic manipulation is straightforward. The 5-mark allocation confirms it's a moderate problem, slightly easier than average due to its predictable structure.
Spec1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1

The first term of a convergent geometric progression is 10. The sum of the first 4 terms of the progression is \(p\) and the sum of the first 8 terms of the progression is \(q\). It is given that \(\frac{q}{p} = \frac{17}{16}\). Find the two possible values of the sum to infinity. [5]

Question 6:
AnswerMarks
6( 1−r8)
10
( 1−r8) ( 1−r4)
1−r 17 17
= [a =  a ]
10 ( 1−r4) 16 (1−r) 16 (1−r)
AnswerMarks Guidance
1−rM1* 17
OE, i.e. substituting p and q expressions into ratio .
16
( 1−r4) ( 1−r8)
16=a , 17= a gets M0 unless recovered later.
(1−r) (1−r)
AnswerMarks Guidance
Simplifying to 16r8−17r4+1 =0  (or equivalent form)DM1 ( 1−r8)
( 1+r4) 17
Or = = .
( 1−r4)
16
( 16r4 −1 )( r4 −1 ) =0 r= 1
 
AnswerMarks Guidance
2A1 1 1
Or r4 =  r= (condone extra r=1solution).
16 2
10
S =
  1 
1−  
AnswerMarks Guidance
 2 DM1 Use of correct sum to infinity formula with either of their r
values providing r 1.
20
S =20 and
AnswerMarks Guidance
3A1 Allow 6.67 or better.
A0 if there is only one or more than two S values.
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 6:
6 | ( 1−r8)
10
( 1−r8) ( 1−r4)
1−r 17 17
= [a =  a ]
10 ( 1−r4) 16 (1−r) 16 (1−r)
1−r | M1* | 17
OE, i.e. substituting p and q expressions into ratio .
16
( 1−r4) ( 1−r8)
16=a , 17= a gets M0 unless recovered later.
(1−r) (1−r)
Simplifying to 16r8−17r4+1 =0  (or equivalent form) | DM1 | ( 1−r8)
( 1+r4) 17
Or = = .
( 1−r4)
16
( 16r4 −1 )( r4 −1 ) =0 r= 1
 
2 | A1 | 1 1
Or r4 =  r= (condone extra r=1solution).
16 2
10
S =
  1 
1−  
 2  | DM1 | Use of correct sum to infinity formula with either of their r
values providing r 1.
20
S =20 and

3 | A1 | Allow 6.67 or better.
A0 if there is only one or more than two S values.

5
Question | Answer | Marks | Guidance
The first term of a convergent geometric progression is 10. The sum of the first 4 terms of the progression is $p$ and the sum of the first 8 terms of the progression is $q$. It is given that $\frac{q}{p} = \frac{17}{16}$.

Find the two possible values of the sum to infinity. [5]

\hfill \mbox{\textit{CAIE P1 2024 Q6 [5]}}