Higher November 2021 Paper 2 Q20
20 Here are the first four terms of an arithmetic series.
\(k \qquad \dfrac{3k}{4} \qquad \dfrac{k}{2} \qquad \dfrac{k}{4}\)
Given that the 15th term of the series is \((90 + 2k)\),
calculate the sum of the first 30 terms of the series.
(5)
| Scheme | Marks |
|---|---|
\(\dfrac{3k}{4} - k\) or \(\dfrac{k}{2} - \dfrac{3k}{4}\) or \(\dfrac{k}{4} - \dfrac{k}{2}\ \left(= -\dfrac{k}{4}\right)\) or \(\dfrac{90 + 2k - k}{14} = \left(\dfrac{90 + k}{14}\right)\) | M1 |
eg \(90 + 2k = k + (15 - 1)\left(\text{‘}{\dfrac{3k}{4} - k}\text{’}\right)\) oe or \(\text{‘}{\dfrac{3k}{4} - k}\text{’} = \text{‘}{\dfrac{90 + k}{14}}\text{’}\) oe | M1 |
| \(k = -20\) | A1 |
| \(\dfrac{30}{2}\left[2(\text{‘}{-20}\text{’}) + (30 - 1)\left(\dfrac{-\text{‘}{-20}\text{’}}{4}\right)\right]\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 1575 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for finding the common difference (\(d\)) in terms of \(k\)
M1: dep equating 2 different expressions in terms of \(k\) using their value(s) of \(d\) in terms of \(k\) (or from working using \(k\))
or other correct method to find \(k\)
M1: dep on previous M1 for correctly substituting, into
\((S_n =)\ \dfrac{30}{2}\left[2k + (30 - 1)d\right]\) or
\(\dfrac{30}{2}(k + l)\) where \(l = k + 29d\)
all values to be numerical