| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| 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 straightforward geometric sequence problem requiring students to equate ratios (forming a quadratic in k) and then apply the sum to infinity formula. The algebra with powers of 3 is routine, and both parts follow standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}} \Rightarrow \dfrac{3^{2(7-2k)}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{3^{2(7-2k)}}\) | M1 | Uses 3 terms to set up equation in \(k\); reaches common base replacing \(9\) with \(3^2\) or \(3\) with \(9^{0.5}\), using power laws correctly |
| \(3^{2(7-2k)-(4k-5)} = 3^{2(k-1)-2(7-2k)}\) leading to \(28-8k = 6k-16 \Rightarrow k=\ldots\) | dM1 | Correct processing leading to a value for \(k\) |
| \(k = \dfrac{5}{2}\)* | A1* | Correct value following correct working. Allow \(k=2.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 3^{4(2.5)-5}\) and \(r = \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}} \Rightarrow\) one of \(a=243\) or \(r=\dfrac{1}{3}\) | M1 | Substitutes their \(k\) (or \(k=\frac{5}{2}\)) to find \(a\) and \(r\) |
| \(S_\infty = \dfrac{a}{1-r} = \dfrac{\text{"243"}}{1-\text{"}\frac{1}{3}\text{"}}\) | M1 | Correct formula \(S_\infty = \dfrac{a}{1-r}\) with their \(a\) and \(r\) |
| \(S_\infty = \dfrac{729}{2}\ (364.5)\) cao | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces expressions for first term and common ratio using \(k = \frac{5}{2}\); finds \(a = 243\) or \(r = \frac{1}{3}\) | M1 | Allow if seen in (a). May be implied by correct values. \(a = 3^{4(2.5)-5}\), \(r = \frac{9^{7-2(2.5)}}{3^{4(2.5)-5}}\) |
| Recalls sum to infinity formula and substitutes values for \(a\) and \(r\) provided \( | r | < 1\) |
| Correct sum to infinity (cao) | A1 | Answer only (no working) scores full marks. Apply isw. |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}} \Rightarrow \dfrac{3^{2(7-2k)}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{3^{2(7-2k)}}$ | M1 | Uses 3 terms to set up equation in $k$; reaches common base replacing $9$ with $3^2$ or $3$ with $9^{0.5}$, using power laws correctly |
| $3^{2(7-2k)-(4k-5)} = 3^{2(k-1)-2(7-2k)}$ leading to $28-8k = 6k-16 \Rightarrow k=\ldots$ | dM1 | Correct processing leading to a value for $k$ |
| $k = \dfrac{5}{2}$* | A1* | Correct value following correct working. Allow $k=2.5$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 3^{4(2.5)-5}$ and $r = \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}} \Rightarrow$ one of $a=243$ or $r=\dfrac{1}{3}$ | M1 | Substitutes their $k$ (or $k=\frac{5}{2}$) to find $a$ and $r$ |
| $S_\infty = \dfrac{a}{1-r} = \dfrac{\text{"243"}}{1-\text{"}\frac{1}{3}\text{"}}$ | M1 | Correct formula $S_\infty = \dfrac{a}{1-r}$ with their $a$ and $r$ |
| $S_\infty = \dfrac{729}{2}\ (364.5)$ cao | A1 | Correct answer |
# Mark Scheme Extraction
---
## Question (Geometric Series - part from first page)
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces expressions for first term **and** common ratio using $k = \frac{5}{2}$; finds $a = 243$ or $r = \frac{1}{3}$ | M1 | Allow if seen in (a). May be implied by correct values. $a = 3^{4(2.5)-5}$, $r = \frac{9^{7-2(2.5)}}{3^{4(2.5)-5}}$ |
| Recalls sum to infinity formula and substitutes values for $a$ and $r$ provided $|r| < 1$ | M1 | Dependent on correct attempt for both $a$ and $r$ using $k=2.5$; allow if unprocessed e.g. $\frac{3^{4(2.5)-5}}{1 - \frac{9^{7-2(2.5)}}{3^{4(2.5)-5}}}$ |
| Correct sum to infinity (cao) | A1 | Answer only (no working) scores full marks. Apply isw. |
---\begin{enumerate}
\item The first 3 terms of a geometric sequence are
\end{enumerate}
$$3 ^ { 4 k - 5 } \quad 9 ^ { 7 - 2 k } \quad 3 ^ { 2 ( k - 1 ) }$$
where $k$ is a constant.\\
(a) Using algebra and making your reasoning clear, prove that $k = \frac { 5 } { 2 }$\\
(b) Hence find the sum to infinity of the geometric sequence.
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q9 [6]}}