June 2024 Paper 1 Q9
9. The first 3 terms of a geometric sequence are
\[3^{4k-5} \qquad 9^{7-2k} \qquad 3^{2(k-1)}\]where \(k\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}} \Rightarrow \dfrac{3^{2(7-2k)}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{3^{2(7-2k)}}\) | M1 | 3.1a |
| \(\left(3^{2(7-2k)}\right)^2 = 3^{4k-5} \times 3^{2(k-1)}\) \(\Rightarrow 28 - 8k = 6k - 7 \Rightarrow k = \ldots\) or \(3^{2(7-2k)-(4k-5)} = 3^{2(k-1)-2(7-2k)}\) \(\Rightarrow 19 - 8k = 6k - 16 \Rightarrow k = \ldots\) | dM1 | 1.1b |
| \(k = \dfrac{5}{2}\,*\) | A1* | 2.1 |
| (3) |
Notes
Special cases:
SC 1: For those that verify rather than prove a SC 100 is awarded for substituting \(k = \dfrac{5}{2}\) into all three terms to correctly obtain 243, 81 and 27 with a statement that this is geometric with \(r = \dfrac{1}{3}\) (or equivalent reason). All statements must be correct.
SC 2: Be aware that e.g. \(\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}} \Rightarrow 81^{2(7-2k)} = 9^{6k-7}\) is an incorrect process (without some indication that they have intentionally squared both sides) that fortuitously leads to the correct answer and may score maximum SC 010.
M1: Uses the 3 terms to set up an equation in \(k\) and
- either reaches a common base by replacing 9 with \(3^2\) or by replacing 3 with \(9^{0.5}\) and uses the power law of indices correctly
- or uses the laws of indices correctly to reach \(9^{14-4k} = 3^{6k-7}\) condoning slips in e.g. expanding brackets.
Writing down e.g. \(2(7-2k) - (4k-5) = 2(k-1) - 2(7-2k)\) is sufficient to imply the M1.
dM1: Correct processing leading to a value for \(k\).
A1*: Correct value following correct working. Allow \(k = 2.5\) in place of \(k = \dfrac{5}{2}\)
Condone missing/invisible brackets provided they are recovered correctly.
Alt 1 Using Base 9:
\[\begin{gathered}\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}} \Rightarrow \dfrac{9^{7-2k}}{9^{2k-2.5}} = \dfrac{9^{k-1}}{9^{7-2k}} \text{ o.e. scores M1}\\\Rightarrow 9^{9.5-4k} = 9^{3k-8} \Rightarrow 9.5 - 4k = 3k - 8 \Rightarrow 7k = 17.5 \Rightarrow k = 2.5\end{gathered}\]Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.
Alt 2 Finding \(r\) in terms of \(k\) and using e.g. \(u_3 = ar^2\):
\[\begin{gathered}r = \dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(7-2k)}}{3^{4k-5}}\left\{= 3^{19-8k}\right\} \textbf{ or } r = \dfrac{3^{2(k-1)}}{9^{7-2k}} = \dfrac{3^{2k-2}}{3^{2(7-2k)}}\left\{= 3^{6k-16}\right\}\\\Rightarrow 3^{4k-5} \times \left(3^{19-8k}\right)^2 = 3^{2k-2} \textbf{ or } \Rightarrow 3^{4k-5} \times \left(3^{6k-16}\right)^2 = 3^{2k-2} \text{ scores M1}\\\Rightarrow 3^{4k-5} \times 3^{2(19-8k)} = 3^{2k-2} \Rightarrow 33 - 12k = 2k - 2 \Rightarrow 14k = 35 \Rightarrow k = 2.5\end{gathered}\]Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.
Alt 3 Using Logs Way 1:
\[\begin{gathered}\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}} \Rightarrow \left(9^{7-2k}\right)^2 = 3^{6k-7} \Rightarrow 9^{14-4k} = 3^{6k-7} \text{ scores M1}\\\Rightarrow (14-4k)\log_{3} 9 = 6k - 7\\ \Rightarrow 2(14-4k) = 6k - 7\\ \Rightarrow k = 2.5\end{gathered}\]Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.
Alt 4 Using Logs Way 2:
\[\begin{gathered}\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}}\\\Rightarrow (7-2k)\log_{3} 9 - (4k-5)\log_{3} 3 = (2k-2)\log_{3} 3 - (7-2k)\log_{3} 9 \text{ scores M1}\\\Rightarrow 2(7-2k) - (4k-5) = 2k - 2 - 2(7-2k)\\ \Rightarrow 19 - 8k = 6k - 16\\ \Rightarrow k = 2.5\end{gathered}\]Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.
Alt 5 Recognising that taking \(\log_{3}\) forms an Arithmetic Sequence:
\[\begin{gathered}\{\log_{3}\}\,u_1 = 4k - 5,\ \{\log_{3}\}\,u_2 = 2(7-2k),\ \{\log_{3}\}\,u_3 = 2(k-1)\\\Rightarrow 2(7-2k) - (4k-5) = 2(k-1) - 2(7-2k) \text{ scores M1}\\ \Rightarrow 19 - 8k = 6k - 16\\ \Rightarrow k = 2.5\end{gathered}\]Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.
There is no need to see any mention of log in this approach.
| Scheme | Marks | AO |
|---|---|---|
| \(a = 3^{4(2.5)-5}\) and \(r = \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}} \Rightarrow\) one of \(a = 243\) or \(r = \dfrac{1}{3}\) | M1 | 2.2a |
| \(S_\infty = \dfrac{a}{1-r} = \dfrac{\text{``}243\text{''}}{1 - \text{``}\frac{1}{3}\text{''}}\) | M1 | 1.1b |
| \(S_\infty = \dfrac{729}{2}\ (364.5)\) cao | A1 | 1.1b |
| (3) | ||
| (6 marks) |
Notes
M1: Deduces expressions for the first term and the common ratio using \(k = \dfrac{5}{2}\) in the correct formulae and finds at least one of \(a = 243\) or \(r = \dfrac{1}{3}\). Allow if seen in (a). May be implied by correct values for \(a\) and \(r\). For reference, \(a = 3^{4(2.5)-5}\) and \(r = \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}}\ \left\{\text{or } r = \dfrac{3^{2(2.5-1)}}{9^{7-2(2.5)}}\right\}\)
M1: Recalls the sum to infinity formula and substitutes their values for \(a\) and \(r\) provided \(|r| \lt 1\)
Dependent on a correct attempt to find both \(a\) and \(r\) using \(k = 2.5\) but allow if neither value is correct or if they are unprocessed e.g. \(\dfrac{3^{4(2.5)-5}}{1 - \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}}}\) scores this mark.
A1: cao. Correct sum to infinity. Answer only (with no working) scores full marks. Apply isw.