Arithmetic and Geometric Sequences and Series

Before this lesson: Arithmetic and exact calculation, Algebraic identities and factorisation

Your goal: Compute terms, state index ranges and explain a finite-sum method.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Arithmetic and Geometric Sequences and Series: the key idea

A sequence is an indexed list. State whether its first term is a₀ or a₁ before using a formula. An arithmetic sequence has a constant difference; a geometric sequence has a constant ratio. Pairing a finite arithmetic sum in reverse order gives n(first+last)/2. For a geometric sum $S=a+ar+\ldots +ar^{n-1}$ , subtract rS from S to get $S=\frac{a(1-r^n)}{1-r}$ when r≠1; if r=1, S=na. Telescoping works when neighbouring terms cancel, but the first and last surviving terms must be kept.

A worked example

Find 1 + 2 + … + 40.

Reverse the same sum: 40+39+…+1. Adding term by term gives 40 pairs each equal to 41. Thus 2S=40×41 and S=820.

Your turn: change one thing

Find 1/(1·2)+1/(2·3)+…+1/(9·10).

Try this on paper before opening the explanation.

Compare your reasoning

Since 1/[k(k+1)]=1/k−1/(k+1), the sum telescopes to 1−1/10=9/10.

Pause and check

A trap to avoid: State the domain, keep exact values and explain why each step is allowed. A correct answer still needs a reason.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove $1+2+4+\ldots +2^{n-1}=2^{n}-1$ for every positive integer n.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Let S be the sum. Then $2S=2+4+\ldots +2^{n}$ . Subtracting S cancels all middle terms and leaves $S=2^{n}-1$ . This is an exact finite subtraction for every positive n, including n=1.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. The sequence 4,7,10,… is arithmetic. Find its fourth term.

Foundation

Hint

Add the common difference.

Answer and reasoning

13. The difference is 3, so the next term is 10+3=13.

2. With a₁=3 and $a_{n}=3\cdot 2^{n-1}$ , find a₃.

Foundation

Hint

At n=3 the exponent is 2.

Answer and reasoning

12. a₃=3×2²=12.

3. Find 1+2+…+10.

Core

Hint

Pair the ends.

Answer and reasoning

55. Ten terms have average (1+10)/2, giving 10×11/2=55.

4. Find 1+3+9+27.

Core

Hint

Add or use a geometric sum.

Answer and reasoning

40. The sum is (3⁴−1)/(3−1)=80/2=40.

5. Which identity produces a telescoping sum?

Stretch

1/[k(k+1)] = 1/k − 1/(k+1)
1/[k(k+1)] = 1/k + 1/(k+1)
1/[k(k+1)] = k − (k+1)
1/[k(k+1)] = 1/k²

Hint

Put the right side over a common denominator.

Answer and reasoning

1/[k(k+1)] = 1/k − 1/(k+1). The difference has numerator (k+1)−k=1 over k(k+1).

6. Why must r=1 be handled separately in the geometric-sum formula?

Stretch

The displayed denominator 1−r would be zero
The sum has no value
There are infinitely many terms
All terms become zero

Hint

Look at the denominator.

Answer and reasoning

The displayed denominator 1−r would be zero. For r=1 every term is a and the finite sum is na; division by 1−r is unavailable.

7. If aₙ=2n+1 for n≥0, what is a₀?

Foundation

Hint

Substitute the starting index.

Answer and reasoning

1. 2·0+1=1.

8. Find 1+2+⋯+20.

Core

Hint

Use n(n+1)/2.

Answer and reasoning

210. 20·21/2=210.

9. In $\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)$ , which terms remain?

Stretch

1+1/(n+1)
1−1/(n+1)
1/n
n−1

Hint

Write out the first few and last terms.

Answer and reasoning

1−1/(n+1). Every intermediate reciprocal cancels, leaving the first positive and last negative terms.

Choose your next step

Continue to Complex numbers: optional bridge. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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Arithmetic and Geometric Sequences and Series: the key idea

A worked example

Find 1 + 2 + … + 40.

Your turn: change one thing

Pause and check

Practise and adjust the level

Write a complete argument

My proof notebook

All nine practice questions

1. The sequence 4,7,10,… is arithmetic. Find its fourth term.

2. With a₁=3 and an=3⋅2n−1a_{n}=3\cdot 2^{n-1}an​=3⋅2n−1, find a₃.

3. Find 1+2+…+10.

4. Find 1+3+9+27.

5. Which identity produces a telescoping sum?

6. Why must r=1 be handled separately in the geometric-sum formula?

7. If aₙ=2n+1 for n≥0, what is a₀?

8. Find 1+2+⋯+20.

9. In ∑k=1n(1k−1k+1)\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)∑k=1n​(k1​−k+11​), which terms remain?

Choose your next step

2. With a₁=3 and $a_{n}=3\cdot 2^{n-1}$ , find a₃.

9. In $\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)$ , which terms remain?