| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Show quadratic equation in n |
| Difficulty | Standard +0.8 This question requires applying the sum formula for arithmetic sequences, algebraic manipulation to derive a quadratic relationship, then solving a system with substitution. Part (a) involves careful expansion and simplification of S₃₆ = (S₆)², which is non-routine. Part (b) requires forming and solving a quadratic equation. The multi-step nature and algebraic complexity elevate this above standard sequence questions, though it remains within typical A-level scope. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_6 = 3(2a+5d) = 6a+15d\) | M1 | Uses \(S_n\) for arithmetic sequence with \(n=6\) or \(n=36\) |
| \(S_{36} = 18(2a+35d) = 36a+630d\) | A1 | Finds correct expressions for \(S_6\) and \(S_{36}\) |
| \(36a + 630d = (6a+15d)^2\) | M1 | Forms equation in \(a\) and \(d\) using their \(S_{36} = (\text{their } S_6)^2\) |
| \(36a+630d = 36a^2+90ad+90ad+225d^2\) | R1 | Expands quadratic and collects like terms to obtain printed answer; only award for completely correct solution with no errors |
| \(4a + 70d = 4a^2 + 20ad + 25d^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a + 5d = 25 \Rightarrow d = \frac{25-a}{5}\) | B1 | Uses \(u_n\) for arithmetic sequence with \(n=6\) |
| \(4a+70\left(\frac{25-a}{5}\right) = 4a^2+20a\left(\frac{25-a}{5}\right)+25\left(\frac{25-a}{5}\right)^2\) | M1 | Eliminates \(a\) or \(d\) using their \(a+5d=25\) and printed result in part (a) to obtain a quadratic in one variable |
| \(350 - 10a = 100a + 625 - 50a + a^2\) | A1 | Obtains correct quadratic equation; need not be simplified |
| \(a^2 + 60a + 275 = 0\) | M1 | Solves their quadratic; \(a=-5\), \(a=-55\) (or \(d=6\), \(d=16\)) |
| \(a = -5,\ a = -55\) (or \(d=6,\ d=16\)) | ||
| \(a = -55\) | A1 | Deduces minimum value \(a=-55\); NMS \(a=-55\) scores 5/5 |
# Question 9:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_6 = 3(2a+5d) = 6a+15d$ | M1 | Uses $S_n$ for arithmetic sequence with $n=6$ or $n=36$ |
| $S_{36} = 18(2a+35d) = 36a+630d$ | A1 | Finds correct expressions for $S_6$ and $S_{36}$ |
| $36a + 630d = (6a+15d)^2$ | M1 | Forms equation in $a$ and $d$ using their $S_{36} = (\text{their } S_6)^2$ |
| $36a+630d = 36a^2+90ad+90ad+225d^2$ | R1 | Expands quadratic and collects like terms to obtain printed answer; only award for completely correct solution with no errors |
| $4a + 70d = 4a^2 + 20ad + 25d^2$ | | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a + 5d = 25 \Rightarrow d = \frac{25-a}{5}$ | B1 | Uses $u_n$ for arithmetic sequence with $n=6$ |
| $4a+70\left(\frac{25-a}{5}\right) = 4a^2+20a\left(\frac{25-a}{5}\right)+25\left(\frac{25-a}{5}\right)^2$ | M1 | Eliminates $a$ or $d$ using their $a+5d=25$ and printed result in part (a) to obtain a quadratic in one variable |
| $350 - 10a = 100a + 625 - 50a + a^2$ | A1 | Obtains correct quadratic equation; need not be simplified |
| $a^2 + 60a + 275 = 0$ | M1 | Solves their quadratic; $a=-5$, $a=-55$ (or $d=6$, $d=16$) |
| $a = -5,\ a = -55$ (or $d=6,\ d=16$) | | |
| $a = -55$ | A1 | Deduces minimum value $a=-55$; NMS $a=-55$ scores 5/5 |9 An arithmetic sequence has first term $a$ and common difference $d$.
The sum of the first 36 terms of the sequence is equal to the square of the sum of the first 6 terms.
9
\begin{enumerate}[label=(\alph*)]
\item Show that $4 a + 70 d = 4 a ^ { 2 } + 20 a d + 25 d ^ { 2 }$
9
\item Given that the sixth term of the sequence is 25 , find the smallest possible value of $a$.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q9 [9]}}