| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Relationship between two GPs |
| Difficulty | Standard +0.3 This question tests standard formulas for AP and GP with straightforward algebraic manipulation. Part (a) involves setting up equations using S_n formulas and solving linear equations. Part (b) requires applying the sum to infinity formula and solving for r, then finding a term. All steps are routine applications of memorized formulas with no novel insight required, making it slightly easier than average. |
| Spec | 1.04h 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{10} = S_{15} - S_{10}\) or \(S_{10} = S_{(11\text{ to }15)}\) | M1 | Either statement seen or implied |
| \(5(2a + 9d)\) oe | B1 | |
| \(7.5(2a + 14d) - 5(2a + 9d)\) or \(\frac{5}{2}[(a+10d)+(a+14d)]\) oe | A1 | |
| \(d = \frac{a}{3}\) AG | A1 | Correct answer from convincing working |
| 4 | Condone starting with \(d = \frac{a}{3}\) and evaluating both summations as 25a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((a + 9d) = 36 + (a + 3d)\) | M1 | Correct use of \(a + (n-1)d\) twice and addition of \(\pm 36\) |
| \(a = 18\) | A1 | |
| 2 | Correct answer www scores 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_\infty = 9 \times S_4\); \(\frac{a}{1-r} = 9\cdot\frac{a(1-r^4)}{1-r}\) or \(9(a + ar + ar^2 + ar^3)\) | B1 | May have 12 in place of \(a\) |
| \(9(1 - r^n) = 1\) where \(n = 3, 4\) or \(5\) | M1 | Correctly deals with \(a\) and correctly eliminates '\(1-r\)' |
| \(r^4 = \frac{8}{9}\) oe | A1 | |
| (\(5^{\text{th}}\) term \(=\)) \(10\frac{2}{3}\) or \(10.7\) | A1 | |
| 4 | Final answer of 10.6 suggests premature approximation – award 3/4 www |
## Question 10(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{10} = S_{15} - S_{10}$ or $S_{10} = S_{(11\text{ to }15)}$ | M1 | Either statement seen or implied |
| $5(2a + 9d)$ oe | B1 | |
| $7.5(2a + 14d) - 5(2a + 9d)$ or $\frac{5}{2}[(a+10d)+(a+14d)]$ oe | A1 | |
| $d = \frac{a}{3}$ **AG** | A1 | Correct answer from convincing working |
| | **4** | Condone starting with $d = \frac{a}{3}$ and evaluating both summations as 25a |
---
## Question 10(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(a + 9d) = 36 + (a + 3d)$ | M1 | Correct use of $a + (n-1)d$ twice and addition of $\pm 36$ |
| $a = 18$ | A1 | |
| | **2** | Correct answer www scores 2/2 |
---
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_\infty = 9 \times S_4$; $\frac{a}{1-r} = 9\cdot\frac{a(1-r^4)}{1-r}$ or $9(a + ar + ar^2 + ar^3)$ | B1 | May have 12 in place of $a$ |
| $9(1 - r^n) = 1$ where $n = 3, 4$ or $5$ | M1 | Correctly deals with $a$ and correctly eliminates '$1-r$' |
| $r^4 = \frac{8}{9}$ oe | A1 | |
| ($5^{\text{th}}$ term $=$) $10\frac{2}{3}$ or $10.7$ | A1 | |
| | **4** | Final answer of 10.6 suggests premature approximation – award 3/4 www |
---10
\begin{enumerate}[label=(\alph*)]
\item In an arithmetic progression, the sum of the first ten terms is equal to the sum of the next five terms. The first term is $a$.
\begin{enumerate}[label=(\roman*)]
\item Show that the common difference of the progression is $\frac { 1 } { 3 } a$.
\item Given that the tenth term is 36 more than the fourth term, find the value of $a$.
\end{enumerate}\item The sum to infinity of a geometric progression is 9 times the sum of the first four terms. Given that the first term is 12 , find the value of the fifth term.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2019 Q10 [10]}}