| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Arithmetic progression with parameters |
| Difficulty | Moderate -0.3 This is a standard arithmetic sequence problem requiring systematic application of well-known formulas (nth term and sum formulas). Part (a) involves algebraic manipulation with parameters, which is slightly more demanding than purely numerical questions, but the methods are routine. Parts (b) and (c) are straightforward substitution once the setup is complete. Overall, slightly easier than average due to its structured, formulaic nature. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a + 4d = 4k\) oe | B1 | Sight of this expression |
| \(S_8 = \frac{1}{2}(8)\left[2a+(8-1)\times d\right] = 20k+16\) oe | B1 | Sight of sum of first 8 terms \(= 20k+16\) |
| \(4[8k-8d+7d] = 20k+16 \Rightarrow d = \ldots\) or \(a+4\left(\frac{5k+4-2a}{7}\right)=4k \Rightarrow a=\ldots\) | M1 | Attempts to solve simultaneously for \(a\) or \(d\) in terms of \(k\) only |
| \((d =)\ 3k-4\) oe or \((a =)\ 16-8k\) * | A1 | Either value from correct working |
| \(a = 4k - 4d = 4k - 4(3k-4) \Rightarrow a=\ldots\) or \(16-8k+8d=24 \Rightarrow d=\ldots\) | M1 | Attempts to solve for the other variable |
| \((a =)\ 16-8k\) and \((d =)\ 3k-4\) oe * | A1* | Both values, no obvious errors. Note: A marks cannot be scored using \(a=16-8k\) without proof in \(S_8\) to find \(d\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(16-8k+8("3k-4") = 24 \Rightarrow k=\ldots\) | M1 | Uses given \(a\) and \(d\) in terms of \(k\) in correct term formula for \(u_9\). Alt: set difference between 5th and 9th terms equal to \(24-4k\) |
| \((k =)\ 2.5\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 16-8\times"2.5"(=-4)\), \(d = "3"\times"2.5"-"4"(=3.5)\) | M1 | Substitutes value of \(k\) to find numerical \(a\) and \(d\) |
| \(S_{20} = \frac{1}{2}(20)\left[2\times("-4")+(20-1)\times"3.5"\right]\) | dM1 | Substitutes numerical \(a\), \(d\) and \(n=20\) into correct formula; dependent on previous M1 |
| \((S_{20} =)\ 585\) | A1 | cao |
# Question 14:
## Part (a)(i)&(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a + 4d = 4k$ oe | B1 | Sight of this expression |
| $S_8 = \frac{1}{2}(8)\left[2a+(8-1)\times d\right] = 20k+16$ oe | B1 | Sight of sum of first 8 terms $= 20k+16$ |
| $4[8k-8d+7d] = 20k+16 \Rightarrow d = \ldots$ or $a+4\left(\frac{5k+4-2a}{7}\right)=4k \Rightarrow a=\ldots$ | M1 | Attempts to solve simultaneously for $a$ or $d$ in terms of $k$ only |
| $(d =)\ 3k-4$ oe or $(a =)\ 16-8k$ * | A1 | Either value from correct working |
| $a = 4k - 4d = 4k - 4(3k-4) \Rightarrow a=\ldots$ or $16-8k+8d=24 \Rightarrow d=\ldots$ | M1 | Attempts to solve for the other variable |
| $(a =)\ 16-8k$ and $(d =)\ 3k-4$ oe * | A1* | Both values, no obvious errors. Note: A marks cannot be scored using $a=16-8k$ without proof in $S_8$ to find $d$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $16-8k+8("3k-4") = 24 \Rightarrow k=\ldots$ | M1 | Uses given $a$ and $d$ in terms of $k$ in correct term formula for $u_9$. Alt: set difference between 5th and 9th terms equal to $24-4k$ |
| $(k =)\ 2.5$ oe | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 16-8\times"2.5"(=-4)$, $d = "3"\times"2.5"-"4"(=3.5)$ | M1 | Substitutes value of $k$ to find numerical $a$ and $d$ |
| $S_{20} = \frac{1}{2}(20)\left[2\times("-4")+(20-1)\times"3.5"\right]$ | dM1 | Substitutes numerical $a$, $d$ and $n=20$ into correct formula; dependent on previous M1 |
| $(S_{20} =)\ 585$ | A1 | cao |
---14. The 5 th term of an arithmetic series is $4 k$, where $k$ is a constant.
The sum of the first 8 terms of this series is $20 k + 16$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find, in terms of $k$, an expression for the common difference of the series.
\item Show that the first term of the series is $16 - 8 k$
Given that the 9th term of the series is 24, find
\end{enumerate}\item the value of $k$,
\item the sum of the first 20 terms.\\
\includegraphics[max width=\textwidth, alt={}, center]{de511cb3-35c7-4225-b459-a136b6304b78-40_2257_54_314_1977}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q14 [11]}}