Example 1 — Arithmetic, working backwards to u₁ and d

An arithmetic sequence has u₄ = 17 and u₉ = 42. Find u₁ and the sum of the first 20 terms.

Two unknowns, so two equations. Write each given term with the nth-term formula:

u₄ = u₁ + 3d = 17
u₉ = u₁ + 8d = 42

Note 3d and 8d, not 4d and 9d — the (n − 1).

Subtract the first from the second. The u₁ disappears:

5d = 25   so   d = 5

Substitute back into either equation:

u₁ + 15 = 17   so   u₁ = 2

Now the sum, using Sₙ = (n/2)(2u₁ + (n − 1)d) with n = 20:

S₂₀ = 10(4 + 19 × 5) = 10(4 + 95) = 10 × 99 = 990

u₁ = 2, S₂₀ = 990.

Why subtract rather than substitute? Because the u₁ terms are identical, subtraction removes one unknown in a single line. Whenever two terms of an arithmetic sequence are given, this is always the fastest route.

Example 2 — Geometric with a negative ratio

A geometric sequence has u₂ = −6 and u₅ = 48. Find r and u₁.

Dividing kills u₁, just as subtracting killed it above:

u₅/u₂ = r³        (from the 2nd term to the 5th is three steps)
48/(−6) = −8 = r³
r = −2

Because the root is odd, there is exactly one real value — no ± to worry about here. Then:

u₂ = u₁ r = −6
u₁(−2) = −6
u₁ = 3

r = −2, u₁ = 3. Check: 3, −6, 12, −24, 48. ✓

The check is not decoration. Writing out four terms takes ten seconds and catches a sign error that would otherwise propagate through every remaining part of the question.

Example 3 — Infinite sum, including the condition

The infinite geometric series 4 + 4x + 4x² + 4x³ + … has a sum of 5. (a) State the values of x for which the sum exists. (b) Find x.

(a) Here u₁ = 4 and r = x. The sum of an infinite geometric series exists precisely when |r| < 1, so:

|x| < 1,  that is  −1 < x < 1

(b) Apply S∞ = u₁/(1 − r):

4/(1 − x) = 5
4 = 5(1 − x)
4 = 5 − 5x
5x = 1
x = 1/5

And 1/5 lies in (−1, 1), so it is valid.

x = 1/5.

Part (a) is a whole mark for one inequality, and students skip it constantly because it feels like it is not really a question. It is. Whenever S∞ appears, the convergence condition is somewhere in the marks.

Example 4 — Logarithms, with a solution that must be rejected

Solve log₂(x) + log₂(x − 2) = 3.

Combine the left side with log x + log y = log(xy):

log₂(x(x − 2)) = 3

Convert to exponential form — log_a b = c means a^c = b:

x(x − 2) = 2³ = 8
x² − 2x − 8 = 0
(x − 4)(x + 2) = 0
x = 4  or  x = −2

Now check both in the original equation. If x = −2, the first term is log₂(−2), which does not exist. Reject it.

x = 4.

The rejection is worth a mark, and it only scores if you write it down. "x = −2 is rejected since log₂(−2) is undefined" is the sentence the scheme wants. An answer of "x = 4" alone, with the second root never mentioned, typically loses that mark.

Example 5 — Binomial, picking out one term

Find the coefficient of x⁵ in the expansion of (2x − 3)⁸.

Do not expand. Write the general term, with a = 2x, b = −3, n = 8:

C(8,r) (2x)^(8−r) (−3)^r

The power of x is 8 − r. We want x⁵, so:

8 − r = 5   giving   r = 3

Substitute r = 3 and evaluate each piece separately — this is where sign and coefficient errors happen, so keep them apart:

C(8,3) = 56
(2x)^5  = 32x⁵          ← the 2 is raised to the 5th too
(−3)³   = −27           ← odd power keeps the minus sign

Multiply:

56 × 32 × (−27) = 56 × (−864) = −48384

The coefficient is −48384.

Note the question asked for the coefficient, not the term. The term is −48384x⁵; the coefficient is the number alone. Answering with the x⁵ attached when the coefficient was asked for is usually condoned, but the reverse — giving a bare number when the term was requested — is not.

Example 6 — A "show that" proof

Show that the sum of any three consecutive integers is divisible by 3.

Begin by defining the general case algebraically. "Any three consecutive integers" means:

Let the integers be n, n + 1 and n + 2, where n ∈ ℤ.

Now transform one side:

n + (n + 1) + (n + 2) = 3n + 3
                      = 3(n + 1)

And close with the reasoning sentence:

Since n ∈ ℤ, (n + 1) ∈ ℤ, so the sum is 3 × an integer,
and is therefore divisible by 3.  ∎

Three separate things earn marks here and students commonly supply only the middle one: the algebraic definition of the general case (n, n+1, n+2 with n an integer), the factorisation, and the closing sentence that interprets 3(n + 1) as divisible by 3. Stopping at 3n + 3 is an incomplete proof — nothing has been said about divisibility.

Testing 4 + 5 + 6 = 15 proves nothing at all. It is one case.

Educerie · original worked examples written against the published IB syllabus structure for Mathematics: analysis and approaches Topic 1, first assessment 2021. All scenarios and values are our own. Sources consulted: none. Last reviewed 5 September 2026.