Arithmetic Sequences: Recognising the Pattern and Finding Any Term

An arithmetic sequence is a sequence of numbers in which the difference between each term and the previous one is always the same constant, called the common difference (d). In the sequence 3, 7, 11, 15, 19, the difference between consecutive terms is always 4: d = 4. In the sequence 100, 85, 70, 55, the difference is always −15: d = −15. The common difference can be positive (increasing sequence), negative (decreasing sequence), or zero (constant sequence, where all terms are equal).

To check whether a sequence is arithmetic, simply compute all the consecutive differences and verify they are equal. If the sequence is 2, 5, 9, 14, the differences are 3, 4, and 5 — they are not equal, so it is not arithmetic. This simple check prevents errors when trying to apply arithmetic-sequence formulas to sequences that do not follow the pattern.

Real-world examples are easy to find. Annual salary raises of a fixed amount produce an arithmetic sequence: if you start at $3,000 and receive a $200 raise every year, your salary forms the sequence 3,000; 3,200; 3,400; 3,600 … with d = 200. Equal instalments on an interest-free loan also form a decreasing arithmetic sequence when you track the remaining balance: each month the balance drops by the same instalment amount.

The general-term formula lets you find any element of the sequence directly, without computing all the preceding ones. The formula is aₙ = a₁ + (n − 1)d, where a₁ is the first term, n is the desired position, and d is the common difference. The reasoning is straightforward: from the first term to the nth term there are (n − 1) additions of d. To reach the second term we add d once; the third, twice; the nth, (n − 1) times.

Let us find the 15th term of the sequence 2, 5, 8, 11, … We identify a₁ = 2, d = 3, and n = 15. Applying the formula: a₁₅ = 2 + (15 − 1) × 3 = 2 + 14 × 3 = 2 + 42 = 44. No need to list the 14 intermediate terms. The formula also works in reverse: if we know that 44 appears in the sequence and want its position, we solve 44 = 2 + (n − 1) × 3, obtaining n = 15.

The inverse form is equally useful for finding the common difference when two terms are known. If we know the 1st term is 5 and the 7th term is 29, we write 29 = 5 + (7 − 1)d, giving 6d = 24 and d = 4. This flexibility makes the general-term formula a versatile tool for any type of arithmetic-sequence problem.

The sum of the first n terms of an arithmetic sequence is Sₙ = n(a₁ + aₙ)/2. This formula comes from an elegant argument attributed to Gauss: write the sequence forwards and then backwards, and add the two versions term by term. Each pair of corresponding terms sums to exactly a₁ + aₙ. Since there are n pairs and each was counted twice, the total is n(a₁ + aₙ)/2.

Example: we want the sum of the first 20 terms of 1, 4, 7, 10, … with a₁ = 1 and d = 3. First we find the 20th term: a₂₀ = 1 + (20 − 1) × 3 = 1 + 57 = 58. Then we apply the sum formula: S₂₀ = 20 × (1 + 58)/2 = 20 × 59/2 = 10 × 59 = 590. Summing 20 terms manually would take a long time; the formula solves it in seconds.