Worked examples
The question
Question: an arithmetic sequence has a₃ = 10 and a₇ = 22.
Find the rule for term n.
a3 = 10, a7 = 22
1 Count the jumps
Term 3 is 10.
Term 7 is 22.
We don't know the terms between.
= 10, ?, ?, ?, 22
In an arithmetic sequence we add the same number d every time.
How many jumps?
7 take away 3: 4 jumps.
arithmetic: + d every time
7 − 3 = 4 jumps
Careful!
We count the jumps, not the terms.
5 terms have only 4 jumps between them.
Careful: 5 terms, but 4 jumps
2 Find d
Start at 10, add d 4 times, and we get 22.
So 10 + 4d = 22.
10 + 4d = 22
We want d by itself.
Two things are stuck to d: times 4 and + 10.
The + 10 is on the outside, so it comes off first.
Then the times 4.
To remove 10, we take away 10.
We do it on both sides.
Why both sides?
The = sign means both sides are the same amount.
If we change only one side, they are not the same any more.
10 + 4d − 10 = 22 − 10
10 and − 10 make zero.
Cross them out.
22 − 10 = 12.
22 − 10 = 12
Now we have 4d = 12.
4d = 12
4d means 4 times d.
We want just d.
To remove times 4, we divide by 4.
We do it on both sides, so they stay equal.
4d4 = 124
4 divided by 4 is 1.
So just d is left.
12 divided by 4 is 3.
124 = 3
So d = 3!
d = 3
3 Find the first term
a₁ is 2 jumps before a₃.
Going back, we take away d each time.
a1 = 10 − 2 × 3
2 times 3 is 6.
10 take away 6 is 4.
a1 = 4
4 Write the rule
The rule from the sequences lesson: the first term, plus (n − 1) times the difference.
an = a1 + (n − 1) × d
Put in a₁ = 4 and d = 3.
an = 4 + (n − 1) × 3
The bracket is times 3.
n times 3 is 3n.
−1 times 3 is −3.
an = 4 + 3n − 3
Plain numbers go together: 4 − 3 is 1.
So the rule is aₙ = 3n + 1.
an = 3n + 1
5 Check
Let's check with the terms we were given.
Put n = 3.
We get 10.
a3 = 3(3) + 1 = 10
And n = 7 gives 22.
Both work!
a7 = 3(7) + 1 = 22
Let's look back.
First: count the jumps.
Then: find d.
Then: find the first term.
Then: write the rule.
Last: check.
▶ Watch it solved
The question
Question: an arithmetic sequence has a₂ = 15 and a₆ = 3.
Find the rule for term n.
a2 = 15, a6 = 3
1 Count the jumps
Term 2 is 15.
Term 6 is 3.
We don't know the terms between.
= 15, ?, ?, ?, 3
In an arithmetic sequence we add the same number d every time.
How many jumps?
6 take away 2: 4 jumps.
arithmetic: + d every time
6 − 2 = 4 jumps
Careful!
We count the jumps, not the terms.
5 terms have only 4 jumps between them.
Careful: 5 terms, but 4 jumps
2 Find d
Start at 15, add d 4 times, and we get 3.
So 15 + 4d = 3.
15 + 4d = 3
We want d by itself.
Two things are stuck to d: times 4 and + 15.
The + 15 is on the outside, so it comes off first.
Then the times 4.
To remove 15, we take away 15.
We do it on both sides.
Why both sides?
The = sign means both sides are the same amount.
If we change only one side, they are not the same any more.
15 + 4d − 15 = 3 − 15
15 and − 15 make zero.
Cross them out.
3 − 15 = −12.
3 − 15 = −12
Now we have 4d = −12.
4d = −12
4d means 4 times d.
We want just d.
To remove times 4, we divide by 4.
We do it on both sides, so they stay equal.
4d4 = −124
4 divided by 4 is 1.
So just d is left.
−12 divided by 4 is −3.
−124 = −3
So d = −3!
d = −3
3 Find the first term
a₁ is 1 jump before a₂.
Going back, we take away d each time.
a1 = 15 − 1 × (−3)
1 times −3 is −3.
15 take away (−3) is 18.
a1 = 18
4 Write the rule
The rule from the sequences lesson: the first term, plus (n − 1) times the difference.
an = a1 + (n − 1) × d
Put in a₁ = 18 and d = −3.
an = 18 + (n − 1) × (−3)
The bracket is times −3.
n times −3 is −3n.
−1 times −3 is 3.
an = 18 − 3n + 3
Plain numbers go together: 18 + 3 is 21.
So the rule is aₙ = −3n + 21.
an = −3n + 21
5 Check
Let's check with the terms we were given.
Put n = 2.
We get 15.
a2 = −3(2) + 21 = 15
And n = 6 gives 3.
Both work!
a6 = −3(6) + 21 = 3
Let's look back.
First: count the jumps.
Then: find d.
Then: find the first term.
Then: write the rule.
Last: check.
▶ Watch it solved
The question
Question: an arithmetic sequence has a₂ = 1 and a₇ = 11.
Find the rule for term n.
a2 = 1, a7 = 11
1 Count the jumps
Term 2 is 1.
Term 7 is 11.
We don't know the terms between.
= 1, ?, ?, ?, ?, 11
In an arithmetic sequence we add the same number d every time.
How many jumps?
7 take away 2: 5 jumps.
arithmetic: + d every time
7 − 2 = 5 jumps
Careful!
We count the jumps, not the terms.
6 terms have only 5 jumps between them.
Careful: 6 terms, but 5 jumps
2 Find d
Start at 1, add d 5 times, and we get 11.
So 1 + 5d = 11.
1 + 5d = 11
We want d by itself.
Two things are stuck to d: times 5 and + 1.
The + 1 is on the outside, so it comes off first.
Then the times 5.
To remove 1, we take away 1.
We do it on both sides.
Why both sides?
The = sign means both sides are the same amount.
If we change only one side, they are not the same any more.
1 + 5d − 1 = 11 − 1
1 and − 1 make zero.
Cross them out.
11 − 1 = 10.
11 − 1 = 10
Now we have 5d = 10.
5d = 10
5d means 5 times d.
We want just d.
To remove times 5, we divide by 5.
We do it on both sides, so they stay equal.
5d5 = 105
5 divided by 5 is 1.
So just d is left.
10 divided by 5 is 2.
105 = 2
So d = 2!
d = 2
3 Find the first term
a₁ is 1 jump before a₂.
Going back, we take away d each time.
a1 = 1 − 1 × 2
1 times 2 is 2.
1 take away 2 is −1.
a1 = −1
4 Write the rule
The rule from the sequences lesson: the first term, plus (n − 1) times the difference.
an = a1 + (n − 1) × d
Put in a₁ = −1 and d = 2.
an = −1 + (n − 1) × 2
The bracket is times 2.
n times 2 is 2n.
−1 times 2 is −2.
an = −1 + 2n − 2
Plain numbers go together: −1 − 2 is −3.
So the rule is aₙ = 2n − 3.
an = 2n − 3
5 Check
Let's check with the terms we were given.
Put n = 2.
We get 1.
a2 = 2(2) − 3 = 1
And n = 7 gives 11.
Both work!
a7 = 2(7) − 3 = 11
Let's look back.
First: count the jumps.
Then: find d.
Then: find the first term.
Then: write the rule.
Last: check.
▶ Watch it solved
