Sequence questions look like puzzles, but they are formula work. Two patterns cover almost every exam item: arithmetic (a fixed amount added each step) and geometric (a fixed factor multiplied each step). Both are described completely by two numbers - a starting value and the step - and once you have those, a later term, a running total and even an infinite sum are a single substitution.
The trap is speed, not difficulty. Under time pressure, candidates add the terms one by one for a question the formula answers in ten seconds.
The model omits patterns that are neither type, such as squares (1, 4, 9, 16) or second differences. These need their own recognition, but they are rare compared with the two standard families.
| Quantity | Formula |
|---|---|
| First term | a |
| Common difference | d = any term − previous term |
| nth term | a + (n − 1)d |
| Sum of n terms | (n ÷ 2) × (2a + (n − 1)d) |
| Sum of n terms with last term l | (n ÷ 2) × (a + l) |
| Quantity | Formula |
|---|---|
| First term | a |
| Common ratio | r = any term ÷ previous term |
| nth term | a × r^(n − 1) |
| Sum of n terms (r ≠ 1) | a × (r^n − 1) ÷ (r − 1) |
| Sum to infinity (|r| < 1) | a ÷ (1 − r) |
Notice the shared skeleton: both nth-term formulas use (n − 1) steps from the first term, because the first term is already at n = 1.
The inverse trick. When a question swaps the positions — "the pth term is q and the qth term is p" — do not find a and d by brute force. Subtract the two equations: a + (p − 1)d = q and a + (q − 1)d = p give (p − q)d = q − p, so d = −1 whenever p ≠ q. Then a = q − (p − 1)d = p + q − 1. The (p + q)th term collapses: a + (p + q − 1)d = (p + q − 1) − (p + q − 1) = 0. The pattern is worth memorising: positions swap, difference is −1, and the term at the sum of the positions is zero.
a = 7 and d = 4. The 20th term is a + (n − 1)d = 7 + 19 × 4 = 7 + 76 = 83.
The sum of the first 20 terms is (n ÷ 2)(2a + (n − 1)d) = 10 × (14 + 76) = 10 × 90 = 900. Check with the last term: 10 × (7 + 83) = 10 × 90 = 900.
a = 3 and r = 2. The 6th term is 3 × 2^5 = 3 × 32 = 96.
The sum of the first 6 terms is 3 × (2^6 − 1) ÷ (2 − 1) = 3 × 63 = 189. Adding directly confirms it: 3 + 6 + 12 + 24 + 48 + 96 = 189.
a = 8 and r = 1/2, and |r| < 1, so the sum is 8 ÷ (1 − 1/2) = 8 ÷ 0.5 = 16.
Write the two equations a + (p − 1)d = q and a + (q − 1)d = p, subtract to get d = −1, then a = p + q − 1. The (p + q)th term is a + (p + q − 1)d = (p + q − 1) − (p + q − 1) = 0. No values of p or q are needed — the answer is always 0 when p ≠ q.
What is the 15th term of the arithmetic sequence 5, 9, 13, ...?
What is the sum of the first 10 even numbers, 2 + 4 + 6 + ... + 20?
What is the 5th term of the geometric sequence 2, 6, 18, ...?
What is the sum to infinity of 12 + 6 + 3 + ...?
In an arithmetic progression the pth term is q and the qth term is p (p ≠ q). What is the (p + q)th term?
Without looking back: write the nth term of an arithmetic sequence and the condition under which a geometric series has a finite sum.
The nth term is a + (n − 1)d; a geometric series converges to a finite sum only when |r| < 1.
In an arithmetic sequence the fixed step is called the common , and in a geometric sequence it is called the common . A geometric series has a finite sum only when the absolute value of the ratio is less than . For the arithmetic sequence 3, 7, 11, ... the 10th term is . If the pth term of an AP is q and the qth term is p, the common difference is and the (p + q)th term is .
In an arithmetic sequence the 3rd term is 10 and the 8th term is 25. Find the first term, the common difference, and the sum of the first 20 terms.
Use the nth-term formula twice. The 3rd term gives a + 2d = 10 and the 8th gives a + 7d = 25.
Subtract the first equation from the second: (a + 7d) − (a + 2d) = 25 − 10, so 5d = 15 and d = 3.
Substitute back: a + 2 × 3 = 10, so a = 4. Check the 8th term: 4 + 7 × 3 = 25, as required.
The sum of the first 20 terms is (20 ÷ 2)(2 × 4 + 19 × 3) = 10 × (8 + 57) = 10 × 65 = 650.
What this exposes: two pieces of information fix both unknowns, because each term is one equation in a and d. That is the general method whenever the formula has two constants.
Sequences describe one quantity changing step by step. The next lesson changes the setting entirely: measurements of flat shapes and solids, where the arithmetic is short but the choice of formula and the units decide the answer.