| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Moderate -0.3 This is a straightforward two-part question on geometric series requiring standard formula manipulation. Part (a) involves simple algebraic rearrangement of S_∞ = a/(1-r) = 3a to find r. Part (b) requires setting up an equation using the general term formula (u_2 - u_4 = ar - ar³ = 16) to find a, then applying the finite sum formula. All techniques are routine and commonly practiced, making this slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3a = \frac{a}{1-r} \Rightarrow r = ...\) | M1 | Cancels all \(a\)'s and attempts to rearrange to find a numerical value for \(r\) |
| \(r = \frac{2}{3}\) | A1* | With no errors and at least one intermediate step after \(3a = \frac{a}{1-r}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(ar - ar^3 = 16\) | B1 | Seen or implied |
| \(\frac{10}{27}a = 16 \Rightarrow a = ...\) | M1 | Proceeds to value for \(a\) from linear equation using \(r = \frac{2}{3}\) and \(ar - ar^3 = 16\) |
| \(a = 43.2\) | A1 | Or any equivalent correct numerical expression e.g. \(\frac{16 \times 27}{10}\), \(\frac{216}{5}\) |
| \(S_{10} = \frac{43.2(1-(\frac{2}{3})^{10})}{1-\frac{2}{3}} = 127.4\) | dM1A1 | dM1: substitutes their \(a\), \(r=\frac{2}{3}\) and \(n=10\) into correct sum formula. A1: awrt 127.4, exact answer is \(\frac{92840}{729}\) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3a = \frac{a}{1-r} \Rightarrow r = ...$ | M1 | Cancels all $a$'s and attempts to rearrange to find a numerical value for $r$ |
| $r = \frac{2}{3}$ | A1* | With no errors and at least one intermediate step after $3a = \frac{a}{1-r}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $ar - ar^3 = 16$ | B1 | Seen or implied |
| $\frac{10}{27}a = 16 \Rightarrow a = ...$ | M1 | Proceeds to value for $a$ from linear equation using $r = \frac{2}{3}$ and $ar - ar^3 = 16$ |
| $a = 43.2$ | A1 | Or any equivalent correct numerical expression e.g. $\frac{16 \times 27}{10}$, $\frac{216}{5}$ |
| $S_{10} = \frac{43.2(1-(\frac{2}{3})^{10})}{1-\frac{2}{3}} = 127.4$ | dM1A1 | dM1: substitutes their $a$, $r=\frac{2}{3}$ and $n=10$ into correct sum formula. A1: awrt 127.4, exact answer is $\frac{92840}{729}$ |
---\begin{enumerate}
\item A geometric sequence has first term $a$ and common ratio $r$
\end{enumerate}
Given that $S _ { \infty } = 3 a$\\
(a) show that $r = \frac { 2 } { 3 }$
Given also that
$$u _ { 2 } - u _ { 4 } = 16$$
where $u _ { k }$ is the $k ^ { \text {th } }$ term of this sequence,\\
(b) find the value of $S _ { 10 }$ giving your answer to one decimal place.
\hfill \mbox{\textit{Edexcel P2 2022 Q8 [7]}}