June 2025 Paper 3 Q4
4 The first term of a geometric sequence is 6.
The fourth term is \(\frac{2}{9}\).
The sequence is infinite.
Find the sum of the associated series. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(6r^3 = \dfrac{2}{9}\) | B1 | 3.1a |
| \(r^3 = \frac{1}{27}\) so \(r = \frac{1}{3}\) | M1 | 1.1 |
| \((S_\infty =)\ 9\) | A1 | 1.1 |
| [3] |
Notes
B1: Forming a correct equation
Or first 4 four terms seen \(6, 2, \frac{2}{3}, \frac{2}{9}\)
soi by correct answer or correct value for \(r\)
M1: Attempt to solve their equation for 4th term of a geometric sequence including taking correct root for their equation
Their equation must be from an attempt to give 4th term of a geometric sequence e.g. \(6r^4 = \frac{2}{9} \to r^4 = \frac{1}{27}\) so \(r = \sqrt[4]{\frac{1}{27}} = 0.438[6\ldots]\)
May be implied by correct answer
Ignore extra values of \(r\) for M1
A1: cao, nfww
Uses \(\frac{a}{1-r}\) to find \(S_\infty = \dfrac{6}{2/3}\), answer must be simplified.
A0 if other values also found unless explicitly rejected