A sequence is a list; a series is its total
A sequence is an ordered list of numbers, each called a term and written a₁, a₂, a₃, … The rule producing them may be given explicitly — "aₙ = 3n − 1" — or recursively, as a starting value plus an instruction for getting from one term to the next.
A series is what you get when you add the terms of a sequence. Sₙ means the sum of the first n terms, and the distinction matters: a question asking for the tenth term and one asking for the sum of the first ten are different questions with different formulae.
| Type | How each term is built | nth term |
|---|---|---|
| Arithmetic (AP) | add a fixed d | aₙ = a + (n − 1)d |
| Geometric (GP) | multiply by a fixed r | aₙ = a rⁿ⁻¹ |
| Harmonic (HP) | the reciprocals form an AP | take reciprocals, use the AP rule, invert back |
Choose Geometric and pull r below 1. The bars shrink towards nothing and the running-total line flattens out — that flattening is the sum to infinity, and it only happens when |r| < 1.
Arithmetic progressions
Everything about an AP follows from one number: the common difference d = aₙ − aₙ₋₁, which must be the same between every consecutive pair. Testing that is how you prove a sequence is arithmetic.
The sum has two equivalent forms, and choosing the right one saves a line. Use the first when you know a and d; use the second when you know the first and last terms.
- a
- the first term
- d
- the common differencecan be negative
- n
- the number of termsa positive integer
Where Sₙ = (n/2)(a + l) comes from
Write the sum forwards and then backwards underneath it and add the two rows. Every column totals a + l, and there are n columns, so twice the sum is n(a + l). The story that Gauss found this at school may or may not be true; the trick certainly works, and it is worth knowing rather than memorising.
Geometric progressions, and when a series can be infinite
A GP has a constant common ratio r = aₙ / aₙ₋₁. Test a sequence by dividing consecutive terms, not by subtracting them.
The most interesting result in the chapter is that an infinite series can have a finite total. If |r| < 1 each term is a fraction of the one before, the terms shrink towards zero fast enough that the running total settles on a limit, and the sum to infinity exists. If |r| ≥ 1 the terms do not shrink and the total grows without bound — there is no sum to infinity, and saying so is the answer.
The third term of a GP is 12 and the sixth is 96. Find a, r, and the sum of the first eight terms.
ar² = 12andar⁵ = 96.Write both given terms using aₙ = arⁿ⁻¹ before doing anything else.- Divide the second by the first:
r³ = 96/12 = 8, sor = 2.Dividing eliminates a in one step. Subtracting would not, which is why division is the move for a GP. - Substitute back:
a(4) = 12, soa = 3. S₈ = 3(1 − 2⁸)/(1 − 2) = 3(1 − 256)/(−1) = 765.With r > 1 it is tidier to use a(rⁿ − 1)/(r − 1), which avoids two negatives.
a = 3, r = 2, S₈ = 765
Means, and the standard trap
Inserting means between two numbers is a standard question. To insert k arithmetic means between a and b, treat the whole thing as an AP with k + 2 terms, first term a and last term b, and find d. For k geometric means, do the same with a GP and find r.
For any two positive numbers, the arithmetic mean is always greater than or equal to the geometric mean, with equality only when the two numbers are the same. That result — AM ≥ GM — is examinable in its own right.
Count the terms, not the means
Inserting three arithmetic means between 2 and 18 gives a sequence of five terms, not three. So 18 = 2 + 4d and d = 4, giving 2, 6, 10, 14, 18. Using n = 3 instead of n = 5 is the single commonest error in this chapter and loses the whole question.
Sigma notation, and the standard sums
Σ is shorthand for "add these up". Σ(from r=1 to n) aᵣ means substitute r = 1, 2, 3 … n into the expression and total the results. Three standard sums are quoted rather than derived, and questions combine them.
Before you leave this chapter
- AP: constant difference, aₙ = a + (n−1)d, Sₙ = (n/2)[2a + (n−1)d].
- GP: constant ratio, aₙ = arⁿ⁻¹, Sₙ = a(1 − rⁿ)/(1 − r).
- S∞ = a/(1 − r) exists only when |r| < 1 — state the condition every time.
- Divide consecutive terms to test for a GP; subtract them to test for an AP.
- Inserting k means makes a sequence of k + 2 terms.
Deciding which progression a word problem is
Applied questions rarely name the progression. The wording does it for you, and the distinction is always the same: does the quantity change by a fixed amount each step, or by a fixed factor?
| The question says | Progression | Because |
|---|---|---|
| salary rises by Rs 2000 each year | AP | a fixed amount is added |
| salary rises by 5% each year | GP | a fixed multiplier of 1.05 |
| a stack with one fewer brick per layer | AP | a constant difference |
| a population doubling every hour | GP | r = 2 |
| a ball rebounding to ¾ of its height | GP | r = 0.75 |
| simple interest | AP | the same interest added each period |
| compound interest | GP | interest on the growing total |
A percentage change is always geometric
An increase of 8% multiplies by 1.08; a decrease of 8% multiplies by 0.92. Both are constant ratios, so the sequence is geometric even though the wording says "increases by". Treating a percentage change as a common difference is the most frequent misreading in the applied questions.
The binomial expansion for a positive integer power
Expanding (a + b)ⁿ by multiplying out is fine for n = 2 and unbearable by n = 7. The binomial theorem writes the answer directly, because the coefficients follow a pattern that can be computed rather than discovered.
Each term is built the same way: the powers of a count down from n, the powers of b count up from 0, and every term has powers summing to n. The coefficient of each term counts how many ways that combination of a's and b's can be chosen from the n brackets — which is exactly what ⁿCr means.
- n
- the powerany positive integer
- r
- which termcounts from 0, so term r+1
- ⁿCr
- the coefficientn! / (r!(n − r)!)
Find the coefficient of x³ in the expansion of (2 + 3x)⁵.
- Identify a = 2, b = 3x, n = 5, and the term needed is r = 3.x³ comes from b³, and b carries the x, so r is 3.
- General term = ⁵C₃ · 2⁵⁻³ · (3x)³.Substituting into the general term rather than writing out all six terms.
- = 10 × 4 × 27x³.⁵C₃ = 10, 2² = 4, and (3x)³ = 27x³ — the 3 is cubed as well as the x, which is the step most often missed.
- = 1080x³, so the coefficient is 1080.The question asks for the coefficient, so the answer is the number alone, without the x³.
1080
Cube the whole term, not just the x
In (3x)³ the bracket means 27x³, not 3x³. Whenever the term in the bracket has a coefficient, raising it to a power raises that coefficient too. Writing the term in brackets before expanding it makes this hard to forget — and it is the single most common error in binomial questions.