| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Form and solve quadratic in parameter |
| Difficulty | Standard +0.3 This is a standard geometric sequence problem requiring students to use the property that consecutive terms have a constant ratio, leading to a quadratic equation. The algebra is straightforward, and part (b) tests understanding of convergence conditions (|r| < 1) and the sum to infinity formula—all routine A-level techniques with no novel insight required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{12-3k}{3k+4} = \frac{k+16}{12-3k}\) | M1 | Forms correct equation linking the three terms; condone invisible brackets if implied by further work |
| \(3k^2 - 62k + 40 = 0\) | A1* | Achieves given quadratic with no errors including invisible brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(3k^2 - 62k + 40 = 0 \Rightarrow k = \ldots\) | M1 | Attempts to solve quadratic achieving at least one value for \(k\) |
| States \(k = 20\) and gives reason e.g. this gives \( | r | < 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(a = 64\) and \(r = -\frac{3}{4}\) (or allow \(a=6\) and \(r=\frac{5}{3}\)) | B1 | May be implied by later correct calculation for \(S_\infty\) |
| \(S_\infty = \frac{``64"}{1-``\left(-\frac{3}{4}\right)"} = \ldots\) | M1 | Full attempt to find \(S_\infty\) using \( |
| \(S_\infty = \frac{256}{7}\) | A1 | \(\frac{256}{7}\) cao; do not allow 36.6 |
## Question 9:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{12-3k}{3k+4} = \frac{k+16}{12-3k}$ | M1 | Forms correct equation linking the three terms; condone invisible brackets if implied by further work |
| $3k^2 - 62k + 40 = 0$ | A1* | Achieves given quadratic with no errors including invisible brackets |
### Part (b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $3k^2 - 62k + 40 = 0 \Rightarrow k = \ldots$ | M1 | Attempts to solve quadratic achieving at least one value for $k$ |
| States $k = 20$ and gives reason e.g. this gives $|r| < 1$ | A1 | e.g. $k=20$ since $|r| = 0.75 < 1$; do not accept reason that just states sequence is converging |
### Part (b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $a = 64$ **and** $r = -\frac{3}{4}$ (or allow $a=6$ and $r=\frac{5}{3}$) | B1 | May be implied by later correct calculation for $S_\infty$ |
| $S_\infty = \frac{``64"}{1-``\left(-\frac{3}{4}\right)"} = \ldots$ | M1 | Full attempt to find $S_\infty$ using $|r|<1$ value of $r$ and corresponding $a$; condone sign slips in $a$ and $r$ only |
| $S_\infty = \frac{256}{7}$ | A1 | $\frac{256}{7}$ cao; do not allow 36.6 |\begin{enumerate}
\item The first three terms of a geometric sequence are
\end{enumerate}
$$3 k + 4 \quad 12 - 3 k \quad k + 16$$
where $k$ is a constant.\\
(a) Show that $k$ satisfies the equation
$$3 k ^ { 2 } - 62 k + 40 = 0$$
Given that the sequence converges,\\
(b) (i) find the value of $k$, giving a reason for your answer,\\
(ii) find the value of $S _ { \infty }$
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q9 [7]}}