Sequences and Series

Sequences are the most formula-dense topic in algebra, and the way to keep it manageable is to notice that almost every formula is a consequence of two ideas: an arithmetic progression adds a fixed amount each step, and a geometric one multiplies by a fixed factor. Everything else — harmonic progressions, arithmetico-geometric series, the means inequality — is built from those two. The one habit that pays here is to identify the type before reaching for a formula, because the commonest error in this chapter is applying an AP formula to a GP or assuming an infinite geometric series converges when it does not.

Arithmetic and Geometric Progressions

The core formulae side by side
Quantity	AP	GP
n-th term	a + (n − 1)d	arⁿ⁻¹
Sum of n terms	(n/2)[2a + (n − 1)d]	a(rⁿ − 1)/(r − 1) for r ≠ 1
Sum to infinity	Diverges unless d = 0	a/(1 − r), only if \|r\| < 1
Middle-term property	Each term is the mean of its neighbours	Each term squared is the product of its neighbours
Useful symmetry	Take terms as a − d, a, a + d	Take terms as a/r, a, ar

🧠 Choose symmetric terms and the unknowns cancel

When a question says "three numbers in AP have sum 15 and product 80", writing them as a − d, a, a + d makes the sum 3a immediately, so a = 5 falls out in one line and only d remains. The same device works for a GP written as a/r, a, ar, where the product becomes a³. For four terms in AP use a − 3d, a − d, a + d, a + 3d so the sum is again free of d. This is not a trick so much as a choice of variables, and it converts simultaneous equations into single ones — which is why it is worth doing even when the algebra looks manageable without it.

⚠️ An infinite geometric series has a sum only when |r| < 1

The formula a/(1 − r) is quoted so often that candidates apply it without checking the ratio, but for |r| ≥ 1 the terms do not shrink and the series has no finite sum at all. So a question giving a series with r = 3/2 and asking for its sum to infinity is asking you to recognise that no such sum exists — and the option list will contain a plausible number computed from the formula anyway. The boundary case r = 1 is worth noting separately: the series becomes a + a + a + … which diverges, and the finite-sum formula a(rⁿ − 1)/(r − 1) also breaks down, so for r = 1 the sum of n terms is simply na.

Means and the AM–GM–HM Inequality

For two positive numbers a and b the arithmetic mean is (a + b)/2, the geometric mean is √(ab) and the harmonic mean is 2ab/(a + b). Two relations follow: G² = A × H, so the geometric mean is itself the geometric mean of the other two, and A ≥ G ≥ H, with equality throughout only when a = b. A harmonic progression is simply a sequence whose reciprocals form an AP, which is why every HP question is best solved by inverting the terms first.

🎯 Why AM ≥ GM is the quickest route to many maximum and minimum problems

The inequality says that for positive numbers the arithmetic mean can never fall below the geometric mean, and equality holds only when all the numbers are equal. That last clause is the useful part, because it tells you where the extremum sits without any calculus. To minimise x + 1/x for x > 0, note that the arithmetic mean of x and 1/x is at least their geometric mean, which is 1, so the sum is at least 2 and equals 2 exactly when x = 1/x, that is x = 1. The same reasoning gives the minimum of a sum whose product is fixed, or the maximum of a product whose sum is fixed — two whole families of questions answered by one inequality, and considerably faster than differentiating.

Special Sums and Telescoping

Standard sums worth memorising
Series	Sum
Σn from 1 to n	n(n + 1)/2
Σn²	n(n + 1)(2n + 1)/6
Σn³	[n(n + 1)/2]² — the square of Σn
Sum of the first n odd numbers	n²
Sum of the first n even numbers	n(n + 1)
Arithmetico-geometric, infinite	a/(1−r) + dr/(1−r)², for \|r\| < 1

🧠 Split the general term into a difference and the series collapses

For a series like Σ 1/(n(n+1)) the direct sum looks hard, but writing the term as 1/n − 1/(n+1) makes almost everything cancel: the sum from 1 to n becomes simply 1 − 1/(n+1). This is telescoping, and it is the standard method for any term that factorises into consecutive pieces — 1/(n(n+2)) splits with a factor of ½, and 1/(n(n+1)(n+2)) splits similarly. The clue that telescoping is intended is a denominator that is a product of terms in AP. Recognising it converts an apparently intractable sum into subtracting the last surviving term from the first, and it is also how the infinite versions are evaluated.

Two closing points. An arithmetico-geometric series has terms of the form (a + nd)rⁿ, and the standard method is to multiply the whole series by r, subtract, and reduce it to a plain GP — the same manoeuvre works for every such series, so it is worth practising once rather than memorising the result. And note that the sum of n terms of a series determines the terms themselves: aₙ = Sₙ − S_n−1, which is how questions that give you Sₙ as a polynomial are meant to be attacked.

Key takeaways

An AP adds a fixed d; a GP multiplies by a fixed r. Identify the type before choosing a formula.
Write three AP terms as a − d, a, a + d and three GP terms as a/r, a, ar — the unknowns cancel.
a/(1 − r) requires |r| < 1; for r ≥ 1 an infinite geometric series has no sum at all.
G² = A × H, and A ≥ G ≥ H with equality only when the numbers are equal.
AM ≥ GM locates extrema without calculus, since equality pins down where they occur.
An HP is a sequence whose reciprocals form an AP — invert the terms first, always.
Σn³ = (Σn)², a coincidence worth remembering because it saves a whole formula.
A denominator that is a product of AP terms signals telescoping — split it into a difference.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The sum to infinity of the series 2 + 3 + 4.5 + 6.75 + … is:
1. −4
2. 8
3. 4
4. The series has no finite sum
Show answer
Answer: D — The series has no finite sum
The common ratio is 3/2, which is greater than 1, so the terms grow without bound and no finite sum exists. Applying a/(1 − r) blindly would give 2/(1 − 1.5) = −4, a negative answer for a series of positive terms — an outcome that should itself signal the formula has been used outside its range of validity. Always check |r| < 1 before quoting the sum to infinity.
For x > 0, the minimum value of x + 4/x is:
1. 2
2. 4
3. 8
4. There is no minimum
Show answer
Answer: B — 4
By AM ≥ GM, (x + 4/x)/2 ≥ √(x · 4/x) = 2, so x + 4/x ≥ 4, with equality when x = 4/x, that is x = 2. The inequality gives both the value and the point at which it occurs, which is why it replaces differentiation here. The answer 2 is the geometric mean itself rather than twice it, which is the usual slip when the factor of 2 from the arithmetic mean is dropped.
If A, G and H are the arithmetic, geometric and harmonic means of two positive numbers, then:
1. A² = GH
2. G² = AH
3. H² = AG
4. A = G = H always
Show answer
Answer: B — G² = AH
Substituting A = (a+b)/2 and H = 2ab/(a+b) gives AH = ab = G², so the geometric mean is the geometric mean of the other two. The three are equal only in the special case a = b; in general A ≥ G ≥ H strictly, which is why the last option fails for any pair of distinct numbers.
The sum of the series 1/(1·2) + 1/(2·3) + … + 1/(n(n+1)) is:
1. n/(n + 1)
2. 1/(n + 1)
3. n(n + 1)/2
4. 1 + 1/(n + 1)
Show answer
Answer: A — n/(n + 1)
Each term splits as 1/k − 1/(k+1), so the sum telescopes to 1 − 1/(n+1) = n/(n+1). The denominator being a product of consecutive integers is the signal to split, and once split almost every term cancels against its neighbour, leaving only the first and the last. Letting n → ∞ then gives 1 for the infinite series, which is a useful check on the form of the answer.
Three numbers in AP have sum 15 and the smallest is 2. The largest is:
1. 5
2. 8
3. 10
4. 13
Show answer
Answer: B — 8
Writing the terms as a − d, a, a + d makes the sum 3a = 15, so the middle term is 5 immediately. With the smallest equal to 2 we get d = 3, so the largest is 5 + 3 = 8. Choosing symmetric terms is what makes the middle value fall out in one step, rather than requiring two simultaneous equations in a and d.
Σn³ from 1 to n equals:
1. n(n + 1)(2n + 1)/6
2. [n(n + 1)/2]²
3. n²(n + 1)/2
4. n(n + 1)/2
Show answer
Answer: B — [n(n + 1)/2]²
The sum of the first n cubes is the square of the sum of the first n integers, a striking identity that means you need to remember only Σn and then square it. The first option is Σn², which is a different sum entirely, and the last is Σn itself — keeping the three straight is worth a moment because questions often combine them in one expression.
If the sum of n terms of a series is Sₙ = n² + 2n, its n-th term is:
1. n² + 2n
2. 2n + 1
3. 2n + 3
4. n + 2
Show answer
Answer: B — 2n + 1
Using aₙ = Sₙ − S_n−1 gives (n² + 2n) − ((n−1)² + 2(n−1)) = 2n + 1. The result is linear in n, which confirms the series is an AP with common difference 2 — a useful consistency check, since a quadratic Sₙ always corresponds to an arithmetic progression. Quoting Sₙ itself as the term is the standard slip here.
If a, b, c are in HP, then which of the following is in AP?
1. a, b, c themselves
2. 1/a, 1/b, 1/c
3. a², b², c²
4. aᵏ, bᵏ, cᵏ for any k
Show answer
Answer: B — 1/a, 1/b, 1/c
A harmonic progression is defined by its reciprocals forming an arithmetic progression, so 1/a, 1/b, 1/c are in AP and satisfy 2/b = 1/a + 1/c. This is why the first move with any HP question should be to invert the terms: the problem then becomes an ordinary AP problem with all the familiar formulae available.