Worked examples
The question
Question: find the sum 2 + 5 + 8 + and so on + 47.
2 + 5 + 8 + ... + 47
1 Find the difference
2 Count the terms
Careful!
(47 − 2)/3 counts the jumps.
There is one more term than jumps.
Careful: count the terms, not the jumps
The rule from the sequences lesson: aₙ = 3n − 1.
The last term, 47, is term number n.
So we solve 3n − 1 = 47.
an = 2 + (n − 1) × 3 = 3n − 1
So 3n − 1 = 47.
3n − 1 = 47
We want n by itself.
Two things are stuck to n: times 3 and − 1.
The − 1 is on the outside, so it comes off first.
Then the times 3.
To remove − 1, we add 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.
3n − 1 + 1 = 47 + 1
− 1 and + 1 make zero.
Cross them out.
47 + 1 = 48.
47 + 1 = 48
Now we have 3n = 48.
3n = 48
3n means 3 times n.
We want just n.
To remove times 3, we divide by 3.
We do it on both sides, so they stay equal.
3n3 = 483
3 divided by 3 is 1.
So just n is left.
48 divided by 3 is 16.
483 = 16
So n = 16!
So there are 16 terms.
n = 16
3 Add in pairs
The pairs rule from the sequences lesson: first plus last makes every pair.
Times the number of terms, over 2.
Sn = n(a1 + an)2
Put in n = 16, the first term 2 and the last term 47.
S16 = 16(2 + 47)2
2 + 47 is 49.
S16 = 16 × 492
16 times 49 is 784.
Half of that is 392.
S16 = 392
4 Check
Let's check the pairs: 2 + 47 is 49.
The second and the second-last: 5 + 44 is 49 too.
2 + 47 = 49, 5 + 44 = 49, ...
16 terms make 8 pairs.
8 times 49 is 392.
It works!
8 pairs: 8 × 49 = 392
Let's look back.
First: find the difference.
Then: count the terms.
Then: add in pairs.
Last: check.
▶ Watch it solved
The question
Question: find the sum 5 + 9 + 13 + and so on + 85.
5 + 9 + 13 + ... + 85
1 Find the difference
2 Count the terms
Careful!
(85 − 5)/4 counts the jumps.
There is one more term than jumps.
Careful: count the terms, not the jumps
The rule from the sequences lesson: aₙ = 4n + 1.
The last term, 85, is term number n.
So we solve 4n + 1 = 85.
an = 5 + (n − 1) × 4 = 4n + 1
So 4n + 1 = 85.
4n + 1 = 85
We want n by itself.
Two things are stuck to n: times 4 and + 1.
The + 1 is on the outside, so it comes off first.
Then the times 4.
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.
4n + 1 − 1 = 85 − 1
+ 1 and − 1 make zero.
Cross them out.
85 − 1 = 84.
85 − 1 = 84
Now we have 4n = 84.
4n = 84
4n means 4 times n.
We want just n.
To remove times 4, we divide by 4.
We do it on both sides, so they stay equal.
4n4 = 844
4 divided by 4 is 1.
So just n is left.
84 divided by 4 is 21.
844 = 21
So n = 21!
So there are 21 terms.
n = 21
3 Add in pairs
The pairs rule from the sequences lesson: first plus last makes every pair.
Times the number of terms, over 2.
Sn = n(a1 + an)2
Put in n = 21, the first term 5 and the last term 85.
S21 = 21(5 + 85)2
5 + 85 is 90.
S21 = 21 × 902
21 times 90 is 1890.
Half of that is 945.
S21 = 945
4 Check
Let's check the pairs: 5 + 85 is 90.
The second and the second-last: 9 + 81 is 90 too.
5 + 85 = 90, 9 + 81 = 90, ...
21 terms make 10 pairs and one term in the middle, 45.
10 times 90, plus 45, is 945.
It works!
10 pairs + the middle 45: 10 × 90 + 45 = 945
Let's look back.
First: find the difference.
Then: count the terms.
Then: add in pairs.
Last: check.
▶ Watch it solved
The question
Question: find the sum −7 − 4 − 1 + and so on + 32.
−7 − 4 − 1 + ... + 32
1 Find the difference
2 Count the terms
Careful!
(32 − (−7)) over 3 counts the jumps.
There is one more term than jumps.
Careful: count the terms, not the jumps
The rule from the sequences lesson: aₙ = 3n − 10.
The last term, 32, is term number n.
So we solve 3n − 10 = 32.
an = −7 + (n − 1) × 3 = 3n − 10
So 3n − 10 = 32.
3n − 10 = 32
We want n by itself.
Two things are stuck to n: times 3 and − 10.
The − 10 is on the outside, so it comes off first.
Then the times 3.
To remove − 10, we add 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.
3n − 10 + 10 = 32 + 10
− 10 and + 10 make zero.
Cross them out.
32 + 10 = 42.
32 + 10 = 42
Now we have 3n = 42.
3n = 42
3n means 3 times n.
We want just n.
To remove times 3, we divide by 3.
We do it on both sides, so they stay equal.
3n3 = 423
3 divided by 3 is 1.
So just n is left.
42 divided by 3 is 14.
423 = 14
So n = 14!
So there are 14 terms.
n = 14
3 Add in pairs
The pairs rule from the sequences lesson: first plus last makes every pair.
Times the number of terms, over 2.
Sn = n(a1 + an)2
Put in n = 14, the first term −7 and the last term 32.
S14 = 14(−7 + 32)2
−7 + 32 is 25.
S14 = 14 × 252
14 times 25 is 350.
Half of that is 175.
S14 = 175
4 Check
Let's check the pairs: −7 + 32 is 25.
The second and the second-last: −4 + 29 is 25 too.
−7 + 32 = 25, −4 + 29 = 25, ...
14 terms make 7 pairs.
7 times 25 is 175.
It works!
7 pairs: 7 × 25 = 175
Let's look back.
First: find the difference.
Then: count the terms.
Then: add in pairs.
Last: check.
▶ Watch it solved
