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Edexcel Pure Paper 1 2024 Question 9
Edexcel Pure Paper 1 2024 Question 9 – Geometric Sequences and Proof
❓ The Question
🧠 Before you start
This question tends to look worse than it actually is.
What usually throws people is the mix — indices and sequences together. If those aren’t fully settled, it starts to feel messy quite quickly.
But underneath it, the idea is straightforward:
If it’s geometric, the ratio is the same each time. That’s all you really need.
✏️ Working
Part (a)
The three terms are:
3^{4k-5}, \quad 9^{7-2k}, \quad 3^{2(k-1)}
Because it’s a geometric sequence:
\frac{9^{7-2k}}{3^{4k-5}} = \frac{3^{2(k-1)}}{9^{7-2k}}
That’s the key step. If that line isn’t there, you don’t really get going.
Now deal with the base.
9 = 3^2
So:
9^{7-2k} = 3^{14-4k}
Substitute everything into base 3:
\frac{3^{14-4k}}{3^{4k-5}} = \frac{3^{2(k-1)}}{3^{14-4k}}
Now simplify each side separately.
Left:
= 3^{(14-4k) – (4k-5)}
= 3^{19 – 8k}
Right:
= 3^{2(k-1) – (14-4k)}
= 3^{2k – 2 – 14 + 4k}
= 3^{6k – 16}
So now:
3^{19 – 8k} = 3^{6k – 16}
Same base, so just equate powers:
19 – 8k = 6k – 16
Solve:
19 + 16 = 6k + 8k
35 = 14k
k = \frac{5}{2}
That’s the whole of part (a). Nothing fancy — just careful handling of indices.
Part (b)
Now you use that value.
Start with the first term:
a = 3^{4k-5}
Substitute k = \frac{5}{2}:
a = 3^{4(\frac{5}{2}) – 5}
= 3^{10 – 5}
= 3^5 = 243
Now the ratio.
Take second ÷ first.
Second term:
9^{7-2k} = 9^{7-5} = 9^2 = 81
So:
r = \frac{81}{243} = \frac{1}{3}
Quick check:
|r| < 1 — so the sum to infinity exists.
Use the formula:
S_\infty = \frac{a}{1-r}
Substitute:
S_\infty = \frac{243}{1 – \frac{1}{3}}
Simplify the denominator:
= \frac{243}{\frac{2}{3}}
So:
= \frac{729}{2}
That’s it.
🎯 Where the Marks Are
Part (a) is quite structured.
- first mark → setting up the ratio
- next step → getting everything into base 3
- final mark → solving for k
If the setup is wrong, it’s hard to recover.
Part (b) is more routine.
- one mark for correct a and r
- one for using the formula
- final mark for the answer
You don’t need anything clever here — just accuracy.
⚠️ What Went Wrong
This question caught quite a few people out.
Not because it was hard, but because of small things.
A common mistake was trying to work with base 9 and base 3 at the same time. That usually leads nowhere.
Another one — and this came up a lot — was getting the index laws wrong. Dividing indices instead of subtracting them.
That completely changes the equation and you don’t really recover from it.
Some students ended up with quadratic equations in k, which is a sign something has gone off earlier.
Part (b) was generally better.
But a few people flipped the ratio and used 3 instead of \frac{1}{3}, which means the sum to infinity formula doesn’t apply.
💡 One Small Tip
If you see indices like this, don’t try to force it.
Change everything to the same base first.
Once that’s done, it becomes a normal algebra problem.
🚀 If This Felt Difficult
If this felt harder than it should, it’s usually down to index laws not being fully secure.
That’s very common.
Working through a few of these with an A level maths tutor helps build that confidence quite quickly.
If you want something more structured, an A level maths revision programme helps tie sequences and indices together so they don’t feel like separate topics.
🔗 Next Steps
👨🏫Author Bio
S. Mahandru is an experienced A Level Maths teacher and founder of Exam.Tips, specialising in exam-focused revision techniques and helping students achieve top grades.
❓ Frequently Asked Questions
📌Why do we use ratios here?
Because in a geometric sequence, consecutive terms always have the same ratio.
📌 Do I have to change everything to base 3?
You don’t have to, but it makes things much easier and avoids mistakes.
📌 What’s the main risk in part (a)?
Getting index laws wrong — especially when subtracting powers.
📌 Why check |r| < 1?
Because without that, the sum to infinity formula isn’t valid.