Lesson Guide
Explanation
A sequence that adds the same amount each time is arithmetic. Seven, eleven, fifteen, nineteen adds four each time, so the common difference is four. To reach the nth term you start at seven and make n minus one jumps of four. That simplifies to four n plus three, so the fiftieth term is two hundred and three. A sequence that multiplies by the same amount each time is geometric. Three, six, twelve, twenty-four doubles each time, so the nth term is three times two to the power n minus one. The tenth term is three times five hundred and twelve, which is one thousand five hundred and thirty-six.
Your Turn
Find the 40th term of 5, 14, 23, 32, … Check: d = 9, so the nth term is 9n − 4 and the 40th term is 356.
Watch Out
The nth term is n − 1 jumps from the first. The 50th term of 7, 11, 15, … is 7 + 49 × 4 = 203, not 7 + 50 × 4 = 207.
Explanation
Here is the trick the young Gauss is said to have used. Write one plus two plus all the way to one hundred, then write the same sum backwards underneath. Each column, one and one hundred, two and ninety-nine, and so on, adds to one hundred and one. There are one hundred columns, but that counts the sum twice, so the sum is one hundred times one hundred and one, halved: five thousand and fifty. The same works for any arithmetic series. Count the terms, multiply by the first plus the last, and halve. Seven plus eleven up to two hundred and three has fifty terms, so the sum is fifty times two hundred and ten, halved, which is five thousand two hundred and fifty.
Your Turn
Add 2 + 4 + 6 + … + 60. Check: 30 terms, so 30 × 62 ÷ 2 = 930.
Watch Out
Counting terms: from 7 to 203 in steps of 4 is (203 − 7) ÷ 4 + 1 = 50 terms. Forgetting the + 1 is the fence-post error.
Explanation
When a pattern repeats, you never need to write it all out. A B C D E repeats in blocks of five, so divide the position by five and look at the remainder. One hundred divided by five is twenty remainder zero. Remainder zero means the hundredth letter finishes a block, so it is E, the last letter. Units digits repeat too. The powers of three end in three, nine, seven, one, then three again, a cycle of four. For three to the power two thousand and twenty-six, divide by four: remainder two, so the units digit is the second in the cycle, nine.
Your Turn
What is the units digit of 7100? Check: 7, 49, 343, 2401 end in 7, 9, 3, 1, a cycle of 4; 100 ÷ 4 = 25 remainder 0, so it is the 4th: 1.
Watch Out
Remainder 0 means the LAST item in the block, not the first. 100 ÷ 5 = 20 remainder 0 gives E, not A.
Explanation
In a growing figure, look at what each new stage adds. A row of matchstick squares starts with four sticks, and each new square shares a stick with its neighbour, so it adds only three. That is an arithmetic sequence: four, seven, ten, and n squares take three n plus one sticks. Some sequences are built from their own earlier terms instead. In the Fibonacci sequence each term is the sum of the two before it: one, one, two, three, five, eight, thirteen. If you know any two neighbouring terms, you can run the rule forwards by adding, or backwards by subtracting.
Your Turn
Each term is the sum of the two before it: 2, 5, 7, 12, … Find the 8th term. Check: 19, 31, 50, 81, so the 8th term is 81.
Watch Out
n squares take 3n + 1 sticks, not 4n: neighbouring squares share a stick. Count what each new square adds.
Explanation
Some questions ask whether something can ever happen. Cut two opposite corners off a chessboard, leaving sixty-two squares. Can thirty-one dominoes cover them? Trying for an hour proves nothing. Instead, look at the colours. Opposite corners are the same colour, so you are left with thirty dark squares and thirty-two light ones. Every domino, wherever you put it, covers one dark and one light square. Thirty-one dominoes would need thirty-one dark squares, and there are only thirty. So the answer is no, for certain. The colour count is an invariant: a fact that no move can change. Totals and odd-or-even are the other common invariants. Quiz and exercises now.
Your Turn
The numbers 1, 2, 3, …, 10 are on a board. A move erases any two and writes their sum. Can you end with the single number 54? Check: each move keeps the total the same; it starts at 55, so you always end with 55. No.
Watch Out
Failing to find a way is not a proof of "No". You need something that never changes (a colour count, a parity, a total) and must show the target breaks it.
📌 Key Points — this lesson in a nutshell
- Arithmetic: nth term = first + (n − 1) × d. Geometric: nth term = first × rn−1. There are n − 1 jumps to the nth term, not n.
- Sum by pairing: an arithmetic series adds to (number of terms) × (first + last) ÷ 2. Count terms with (last − first) ÷ d + 1.
- Repeating patterns: divide by the length of the block and use the remainder. Remainder 0 means the last item of the block.
- Growing figures: count what each new piece adds. Shared sticks and corners must not be counted twice.
- Invariants: to prove something can never happen, find a quantity that never changes (a total, a parity, a colour count) and show the target breaks it.
Step 1: Introduction
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