Standard +0.3 This is a straightforward arithmetic series problem requiring finding the common difference from three terms, then applying the sum formula. It involves more algebraic manipulation than typical AS pure questions but follows a standard method with no novel insight required, making it slightly easier than average.
The first three terms of an arithmetic series are \(9p\), \(8p - 3\), \(5p\) respectively, where \(p\) is a constant.
Given that the sum of the first \(n\) terms of this series is \(-1512\), find the value of \(n\). [6]
Expand and attempt to solve 3-term quadratic equation in \(n\)
\(n = 28 \text{ only}\)
A1
This mark should be withheld if \(n = -18\) appears as part of the final answer
$(8p - 3) - 9p = 5p - (8p - 3)$ | M1 | Setting up an equation to find $p$ | Allow a single sign error
$p = 3$ | A1 |
$a = 27, d = -6$ | A1FT | Using their value of $p$ to calculate $a$ and $d$
$\frac{n}{2}[2(27) + (n-1)(-6)] = -1512$ | M1 | Setting up an equation using the correct formula for the sum of an AP equated to $-1512$
$n^2 - 10n - 504 = 0 \Rightarrow (n - 28)(n + 18) = 0$ | M1 | Expand and attempt to solve 3-term quadratic equation in $n$ | Solving of 3-term quadratic may be done BC
$n = 28 \text{ only}$ | A1 | This mark should be withheld if $n = -18$ appears as part of the final answer
The first three terms of an arithmetic series are $9p$, $8p - 3$, $5p$ respectively, where $p$ is a constant.
Given that the sum of the first $n$ terms of this series is $-1512$, find the value of $n$. [6]
\hfill \mbox{\textit{OCR H240/03 2018 Q4 [6]}}