| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Moderate -0.3 Part (a) is a standard geometric progression problem requiring finding the common ratio from consecutive terms, then the first term, and applying the sum to infinity formula - routine multi-step application of formulas. Part (b) is a straightforward arithmetic progression problem using the constraint that angles sum to 360° and the largest is 4 times the smallest - algebraic manipulation but no novel insight required. Both parts are typical textbook exercises slightly easier than average A-level questions due to being from P1 (AS-level content). |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks |
|---|---|
| 7 (a) | 1 2 |
| Answer | Marks |
|---|---|
| 3 | M1 |
| Answer | Marks |
|---|---|
| [4] | Any valid method, seen or implied. |
| Answer | Marks |
|---|---|
| (b) | 4a =a+4d → 3a = 4d |
| Answer | Marks |
|---|---|
| Largest = a + 4d or 4a = 115.2º aef | B1 |
| Answer | Marks |
|---|---|
| [4] | May be implied in |
| Answer | Marks | Guidance |
|---|---|---|
| Page 7 | Mark Scheme | Syllabus |
| Cambridge International AS/A Level – May/June 2015 | 9709 | 11 |
| Answer | Marks |
|---|---|
| (iv) | 1 |
| Answer | Marks |
|---|---|
| x = 2cos −1 3 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 8 | Mark Scheme | Syllabus |
| Cambridge International AS/A Level – May/June 2015 | 9709 | 11 |
Question 7:
--- 7 (a) ---
7 (a) | 1 2
ar² = , ar³ =
3 9
2
→ r = aef
3
3
Substituting → a =
4
3
→ S ∞ = 1 4 = 21 4 aef
3 | M1
A1
M1 A1
[4] | Any valid method, seen or implied.
Could be answers only.
Both a and r
Correct formula with r <1, cao
(b) | 4a =a+4d → 3a = 4d
5
360 = S 5 = (2a+4d) or 12.5a
2
→ a = 28.8º aef
Largest = a + 4d or 4a = 115.2º aef | B1
M1
A1
B1
[4] | May be implied in
360=5/2(a+4a)
Correct S n formula or sum of 5
terms
cao, may be implied
(may use degrees or radians)
Page 7 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 11
8
(i)
(ii)
(iii)
(iv) | 1
f : x⟼5 + 3cos x for 0 ø x ø 2π.
2
1
x
5 + 3cos = 7
2
1 2
x
cos =
2 3
1
x = 0.84 x = 1.68 only, aef
2
(in given range)
2
No turning point on graph or 1:1
1
y = 5 + 3cos x
2
−1
Order; −5, ÷3, cos , ×2
x−5
x = 2cos −1 3 | B1
M1A1
[3]
B1
B1
[2]
B1
[1]
M1
M1
A1
[3] | 1 2
Makes cos x=
2 3
−1
Looks up cos first, then ×2
y always +ve, m always –ve.
from (0, 8) to (2π, 2) (may be
implied)
cao, independent of graph in (ii)
Tries to make x subject.
Correct order of operations
cao
Page 8 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 11\begin{enumerate}[label=(\alph*)]
\item The third and fourth terms of a geometric progression are $\frac{1}{4}$ and $\frac{2}{9}$ respectively. Find the sum to infinity of the progression. [4]
\item A circle is divided into 5 sectors in such a way that the angles of the sectors are in arithmetic progression. Given that the angle of the largest sector is 4 times the angle of the smallest sector, find the angle of the largest sector. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2015 Q7 [8]}}