June 2024 Paper 2 Q7
7 Two arithmetic progressions, \(A\) and \(B\), each have 100 terms denoted by \(a_i\) and \(b_i\) respectively, where \(i = 1, 2, 3, \ldots 100\).
The common difference of \(A\) is \(d\), where \(d\) is a positive integer.
The two progressions have the following properties.
- \(a_1 = b_{100} = 4\)
- \(b_1 = a_{100}\)
Show that, in this case,\[i = \frac{101}{2} - \frac{5}{d}.\] [6]
| Scheme | Marks | AO |
|---|---|---|
| common difference of \(B\) is \(-d\) | B1 | 1.1 |
| \((b_1 = a_{100} =)\ 4 + 99d\) | B1 | 1.1 |
| \(b_i = 4 + 99d + (i - 1)(-d)\quad\) or \(4 + 100d - id\) | M1 | 3.1a |
| Fully correct expression for \(b_i\) | A1 | 2.1 |
| \(4 + 100d - id = 4 + (i - 1)d + 10\) | M1 | 1.1 |
| \((2id = 101d - 10)\) \(\Rightarrow i = \frac{101}{2} - \frac{5}{d}\quad\) AG | A1 | 1.1 |
| [6] |
Notes
B1: soi
B1: soi (or \(a_1 + 99d\))
M1: For an attempt at the general term \(b_i\) using \(b_1\ (= a_{100})\), i.e. an expression of the form \(a_1 + kd + (i - 1)(\pm d)\) oe
M1: Equating their \(b_i\) to an expression for \(a_i + 10\)
May see \(a_1 + (i - 1)d + 10\)
A1: www (so this mark is dependent on all previous marks)
May see \(a_1\) used throughout instead of 4
Condone other letters used throughout (e.g. \(n\) for \(i\) and \(c\) for \(-d\)):
- Must recover to \(d\) in order to solve (A1A1)
- Must recover to \(i\) to demonstrate the AG (A1)
| Scheme | Marks | AO |
|---|---|---|
| \(\frac{101}{2} - \frac{5}{d}\) not always integer [between 1 & 100] | B1 | 2.4 |
| \(d = 2\) or 10 | B1 | 3.2b |
| [2] |
Notes
B1: soi (accept e.g. “\(i\) must be an integer”)
B1: Ignore negative values (but B0 for any additional positive values)