Mark scheme · Mathematics: analysis and approaches

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Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 1 — Topic test · mark scheme

M = method, awarded for a correct approach even if the arithmetic that follows is wrong. A = accuracy, dependent on the method mark above it. R = reasoning, awarded for justification in words.

Follow through (FT) applies throughout: a value carried correctly from an earlier incorrect part still earns the method marks of the later part.

1Arithmetic sequence — 5 marks

(a) [2]

Sets up using `uₙ = u₁ + (n − 1)d`, e.g. `u₁ + 2d = 11` and `u₁ + 7d = 31`, or `5d = 31 − 11`	M1
`d = 4`	A1

Accept the direct route d = (u₈ − u₃)/5. Award M1 for dividing by 5; dividing by 6 is the expected error and scores M0.

(b) [1]

`u₁ = 3`	A1 FT from their `d`

(c) [2]

Substitutes into `Sₙ = (n/2)(2u₁ + (n − 1)d)` with `n = 15`, e.g. `(15/2)(6 + 14 × 4)`	M1
`S₁₅ = 465`	A1

(15/2)(u₁ + u₁₅) = (15/2)(3 + 59) is equally acceptable for M1.

2Geometric sequence — 6 marks

(a) [2]

Forms `u₁r³ = 3`, i.e. `24r³ = 3`, or `r³ = 3/24`	M1
`r³ = 1/8` leading to `r = 1/2`	A1 AG

AG — answer given. The final line must be derived. A response that begins by substituting r = 1/2 and verifying u₄ = 3 scores 0: it assumes what was to be shown. Note r³, not r⁴ — three steps from the 1st term to the 4th.

(b) [2]

Substitutes into `Sₙ = u₁(1 − rⁿ)/(1 − r)`, e.g. `24(1 − (1/2)⁶)/(1 − 1/2)`	M1
`S₆ = 189/4`	A1

Exact form required. 47.25 scores M1 A0. Accept 47¼.

(c) [2]

States the condition is met, e.g. "`\|r\| = 1/2 < 1`, so the series converges"	R1
`S∞ = 24/(1 − 1/2) = 48`	A1

R1 requires the comparison to be stated, not merely implied by computing the sum. "Because r is a fraction" is insufficient — it must reference |r| < 1.

3Logarithmic equation — 4 marks

Combines the logs: `log₅(x(x − 4)) = 1`	M1
Converts to `x(x − 4) = 5`	A1
Solves `x² − 4x − 5 = 0` to obtain `x = 5` and `x = −1`	A1
Rejects `x = −1` with a reason, e.g. "`log₅(−1)` is undefined"	R1

The final mark is for the rejection and its justification. x = 5 given alone, with x = −1 never mentioned, scores a maximum of 3. Rejecting without a reason also scores R0.

4Binomial expansion — 5 marks

Writes a general term, e.g. `C(6,r)(3x)^(6−r)(−2)^r`	M1
Identifies `r = 2` from `6 − r = 4`	A1
Correct components: `C(6,2) = 15`, `(3x)⁴ = 81x⁴`, `(−2)² = 4`	A1
Evaluates `15 × 81 × 4`	M1
Term is `4860x⁴`	A1

Common errors and their consequences: - (3x)⁴ written as 3x⁴ — the coefficient not raised. Loses the third A1, but the final M1 is still available on follow through. - r = 4 used instead of r = 2 (confusing the required power of x with r). A0, but M1 for the general term stands. - Answer given as 4860 when the term was asked for: award full marks only if x⁴ appears somewhere in the working; otherwise deduct the final A1.

Mark distribution

Split	Allocation
By topic	Sequences and series 11 · Logarithms 4 · Binomial theorem 5
By objective	AO1 knowledge and routine technique 10 · AO2 problem solving 6 · AO3 communication and reasoning 4
By command word	Find 12 · Solve 4 · Show that 2 · Explain 2

Total: 20 marks

Topic 1 also assesses simple deductive proof, which this test does not sample. When the proof variant is issued, it replaces question 4 and carries its own 4-mark scheme: M1 for defining the general case algebraically, A1 for forming the expression, A1 for the factorisation, R1 for the concluding sentence that interprets the factorised form. Numerical verification scores zero.

Educerie · original mark scheme for original questions, written against the published IB syllabus structure for Mathematics: analysis and approaches Topic 1, first assessment 2021. Mark-type conventions (M/A/R, AG, FT) follow standard IB marking practice, which is a convention rather than protected expression. Sources consulted: none. Last reviewed 5 September 2026.