Higher June 2025 Paper 1 Q24
24 An arithmetic series has 30 terms.
The first term is \(a\)
The common difference is \(d\)
The 20th term is 123
The sum of the 30 terms is 2880
Work out the value of \(a\) and the value of \(d\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
| \(123 = a + (20 - 1)d\) or \(123 = a + 19d\) | M1 |
\(2880 = \dfrac{30}{2}\left(2a + (30 - 1)d\right)\) or \(2880 = \dfrac{30}{2}\left(2a + 29d\right)\) or \(192 = 2a + 29d\) | M1 |
eg \(192 = 2a + 29d\) \(123 = a + 19d\) (× 2) \(246 = 2a + 38d\) Subtracting \(54 = 9d\) or \(192 = 2(123 - 19d) + 29d\) oe or \(d = 6\) or eg \(192 = 2a + 29d\) (× 19) \(123 = a + 19d\) (× 29) \(3648 = 38a + 551d\) \(3567 = 29a + 551d\) Subtracting \(81 = 9a\) or \(192 = 2a + 29\left(\dfrac{123 - a}{19}\right)\) oe or \(a = 9\) | M1 |
| eg \(192 = 2a + 29(\text{``}{6}\text{''})\) oe or \(123 = a + 19(\text{``}{6}\text{''})\) oe or eg \(192 = 2(\text{``}{9}\text{''}) + 29d\) oe or \(123 = \text{``}{9}\text{''} + 19d\) oe | M1 |
| Working required Answer: \(a = 9\) \(d = 6\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for using \(U_n = a + (n - 1)d\)
M1: for using \(S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)\)
M1: (dep on M2) for a correct method to find \(a\) or \(d\):
coefficients of \(a\) or \(d\) the same in correct equations and correct operator to eliminate selected variable resulting in an equation in \(a\) only or in \(d\) only
or
writing \(a\) or \(d\) in terms of the other variable and correctly substituting (condone missing brackets)
M1: (dep on M3) for substituting their found value of \(a\) or \(d\) into a correct equation
A1: dep on M2
\(a\) and \(d\) must be clearly identified