Let us look at the growing pattern of squares given in the figure. The first four stages of the pattern are shown. If we count the number of tiny squares at each stage, we get sequence 1, 5, 9, 13. Can you predict the number of squares in Stages 5 and 6 of the sequence? In Stages 10, 11 and 12? In Stage 20? At any stage?
Step-by-step solution
Idea: Each stage adds 4 squares, one at the end of each arm. So stage n has the first square plus (n − 1) lots of 4.
- From one stage to the next, one square is added at each of the 4 corners: 1, 1 + 4, 1 + 4 + 4, …½ mark
- Stage 5: 13 + 4 = 17; Stage 6: 17 + 4 = 21.½ mark
- Stage n = 1 + (n − 1) × 4 = 4n − 3.1 mark
- Stage 10: 40 − 3 = 37; Stage 11: 44 − 3 = 41; Stage 12: 48 − 3 = 45.½ mark
- Stage 20: 4 × 20 − 3 = 77.½ mark
Check: 4 × 4 − 3 = 13, matching the 4th stage ✓.
Answer to write in the exam
Stage 1: 1; each stage adds 4 squares
Stage 5 = 13 + 4 = 17, Stage 6 = 17 + 4 = 21
Stage n = 1 + (n − 1) × 4 = 4n − 3
Stage 10 = 37, Stage 11 = 41, Stage 12 = 45
Stage 20 = 4 × 20 − 3 = 77
∴ 17, 21; 37, 41, 45; 77; 4n − 3
Common mistakes that cost marks
- Writing stage n as 4n (giving 4, 8, 12, …). The first stage has 1 square, so subtract 3: 4n − 3.
- Using 1 + 4n instead of 1 + 4(n − 1). Stage 1 has had no squares added yet.
- Counting stage 20 by doubling stage 10 (2 × 37 = 74). The pattern adds a fixed 4 each time; it does not double.
How this can come in the exam
A pattern of tiles has 1, 5, 9, 13, … tiles at stages 1, 2, 3, 4, …. At which stage are there 101 tiles?
- 25
- 26
- 24
- 27
Show answer
(B) 26
4n − 3 = 101 ⇒ 4n = 104 ⇒ n = 26.
Try one yourself
A matchstick pattern uses 4, 7, 10, 13, … matchsticks at stages 1, 2, 3, 4. How many are needed at stage 15 and at stage n?
Show answer
Add 3 each stage: stage n = 4 + 3(n − 1) = 3n + 1; stage 15 needs 46.
More questions like this
- Consider all the sequences we have discussed so far. Which ones are arithmetic progressions and which ones are not? Can you justify your claim?
- Verify that the following sequences are arithmetic progressions and write their nth terms. What do you observe when you plot the ordered pairs emerging from them?
- Using the formula tn = a + (n − 1) × d, find the nth term of the following arithmetic progressions.
- Consider once again the AP: 1, 5, 9, 13, 17, … Since every term is 4 more than the previous term, another way of writing this sequence is t1 = 1, tn = tn−1 + 4 for n ≥ 2. Verify for yourself that this recursive rule leads to the terms of the sequence.
- Find recursive rules for the APs in the previous exercises.
All Sequences and progressions questions · All maths questions