Natural numbers in a row form sequences. The simplest of these, the arithmetic progression, appears in many problems. Move the figure to see a PA grow:
A visual map of progressions
Your mission: transform patterns into arguments
Discoveries itinerary
- Observe a sequence and identify how each term is obtained.
- Recognize a PA by the constant difference between consecutive terms.
- Relate the position of a term to its value using the general term.
- Justify the sum with pairs of terms and apply the idea in a concrete situation.
Start with introduction to progressions and proceed to arithmetic progression.
Connection with naturals: the positions of the terms are 1, 2, 3, … . The values of a PA can also be negative or fractional; not all PAs have only natural terms. In the explorer, watch what happens when you change the ratio.
Challenge of the ranks
An auditorium has 12 rows. The first has 8 chairs and each following row has 2 more chairs. How many chairs are in the last row? What is the total capacity?
Check the answer and strategy
The last row has 8 + (12 − 1) × 2 = 30 chairs. The sum is 12 × (8 + 30) ÷ 2 = 228 seats. When joining the first row with the last row, each pair has 38 chairs; are 6 even.
Explain before using the formula
Why do we use n − 1 in the general term aₙ = a₁ + (n − 1)r?
See a justification
The first term is already in position 1. To get to position n, we take n − 1 steps, each one adding the ratio r.
Ready to complete? Create a PA of natural terms, calculate its tenth term and justify the sum of the first ten terms. Then, investigate what changes with zero or negative ratio.
Laboratory: recognize, calculate and check
In the training below, three missions appear: discover the reason, calculate a term and add terms. Write down your strategy before checking. The difficulty changes the number of terms considered. The questions in this activity use natural terms and ratios, including zero.
Challenge: when do terms stop being natural?
A PA starts at 9 and has reason −2. Write the first six terms. Up to what position are all terms still natural?
Check out the investigation
The terms are 9, 7, 5, 3, 1 and −1. The first five are natural; the sixth is already negative. The position remains a positive natural, even when the value of the term is not natural.