Chapter 2 · Sequences · Concept 4
2-19 Group Sequences
Take the even numbers and chop them into bunches: one, then three, then five. Nothing about the numbers has changed — but now every one of them has two addresses, and almost every question in this lesson is really about translating between them.
Guiding question
How do you find a term when the sequence has been cut into groups?
Learning outcome. Solve problems by dividing a sequence into groups, to find a term or the sum of a group.
Two sequences, not one
For a grouped sequence, look at the sequence before grouping and at how many terms are in each group — and treat them as two separate sequences.
2 | 4, 6, 8 | 10, 12, 14, 16, 18 | 20, 22, …
before grouping: 2, 4, 6, 8, … so aₙ = 2n
group sizes: 1, 3, 5, … so bₙ = 2n − 1
The one fact everything hangs on:
the last term of group n sits at position
b₁ + b₂ + … + bₙ
— the group sizes added up. Once you can write that down, every question here is arithmetic. And the first term of group n is one place after the last term of group n − 1.
The small grey number under each term is where it sits in the whole sequence. The row it is on is which group it belongs to. Keeping those two apart is the entire skill.
Warm Up · before grouping aₙ = 2n, group sizes bₙ = 2n − 1 · click any number
Every number here has two addresses. Click one and see both.
88 is the 44th term overall, which puts it 8th in group 7 — and the second address is what these questions ask for.
Find the first term of group 10
Add the sizes of groups 1 to 9. The next position is where group 10 starts.
sizes → position → value
Find the sum of group n
You need three things: its first term, its last term, and how many it holds. Then it is an ordinary lesson 2-4 sum.
first · last · count
Which group is 88 in?
First find its position in the whole sequence. Then see which group that position falls inside.
value → position → group
All three run through position in the whole sequence. That is the bridge between the two addresses, and every question crosses it at some point — so write the position down explicitly rather than trying to jump straight to the answer.
1. Find the first term of the 10th group.
- 1
Write down the two sequences before anything else.
aₙ = 2n (before grouping) · bₙ = 2n − 1 (group sizes)Sizes are 1, 3, 5, 7 — arithmetic, first 1, difference 2.
2. Find the sum of the n-th group.
- 1
Find where group n ENDS. Add the first n group sizes.
½n{2 × 1 + (n − 1) × 2} = n2A tidy result worth remembering: with sizes 1, 3, 5, … group n ends at position n².
Work with a partner. One of you tracks positions in the whole sequence, the other tracks groups. Say which address you are using every time you write a number down.
1. For 1 | 2, 3 | 4, 5, 6 | 7, 8, 9, 10 | 11, 12, …
2. For 1, 4 | 7, 10, 13, 16 | 19, 22, 25, 28, 31, 34 | 37, 40, …
Four parts, each one building on the last.
Faced with an unfamiliar sequence of real data, what would make you reach for the difference-sequence method of lesson 2-18 rather than the grouping method of this one?
Ask what each method needs before it can start.
Confusing the two addresses.
"The 8th term of the 7th group" and "the 44th term" are the same number described two ways. Say which one you mean every time.
Starting group n at the position where group n − 1 ended.
If group 9 ends at position 81, group 10 starts at 82. The +1 is easy to drop.
Summing the group sizes to n when you wanted n − 1.
To find where group n STARTS, add the sizes up to n − 1. Adding to n tells you where it ends.
Forgetting that the sum of a group needs three things.
First term, last term, and how many. Missing any one of them and the lesson 2-4 formula cannot be used.
Adding group sums one by one when the total is asked for.
Groups 1 to 12 together are just the first 90 terms. One sum, not twelve.
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