Sequences and Series — H2 Maths help

Sequences and series form a core strand of H2 Mathematics. You will meet arithmetic and geometric progressions in both pure algebra and applied modelling contexts, and — since the 2025 revision of 9758 — sequences generated by recurrence relations $u_{n+1} = f(u_n)$ , explored with the graphing calculator. (The method of differences was removed from the syllabus in the same revision, so telescoping-sum questions in older prelim papers are no longer examinable.)

What you need to know

Arithmetic progression (AP): $n$ th term $T_n = a + (n-1)d$ ; sum of first $n$ terms $S_n = \dfrac{n}{2}(2a + (n-1)d)$ .
Geometric progression (GP): $n$ th term $T_n = ar^{n-1}$ ; sum $S_n = \dfrac{a(1-r^n)}{1-r}$ for $r \neq 1$ .
Sum to infinity of a GP: $S_\infty = \dfrac{a}{1-r}$ exists only when $|r| < 1$ .
None of the AP/GP formulas are in the exam formula booklet — see what’s in MF27. They must be memorised.
Sigma notation: $\displaystyle\sum_{r=1}^{n} f(r)$ denotes the sum of $f(r)$ for $r = 1, 2, \ldots, n$ ; know how $S_n$ and the term $u_n = S_n - S_{n-1}$ are related.
Recurrence relations: a sequence can be generated by $u_{n+1} = f(u_n)$ from a starting value $u_1$ . Use the GC’s table or sequence mode to explore behaviour. If the sequence converges to a limit $L$ , then letting $n \to \infty$ in the recurrence gives $L = f(L)$ .
Convergence of a GP series requires $|r| < 1$ ; always state this condition explicitly in exam answers.

Worked example 1 — Sum to infinity of a geometric series

Problem. A GP has first term $8$ and common ratio $r$ . The sum to infinity is $24$ . Find $r$ and the sum of the first 4 terms.

Solution.

Using the sum-to-infinity formula:

S_\infty = \frac{a}{1-r} = 24.

Substituting $a = 8$ :

\frac{8}{1-r} = 24 \implies 1 - r = \frac{1}{3} \implies r = \frac{2}{3}.

Since $|r| = \tfrac{2}{3} < 1$ , the series converges — consistent with the question. Now apply the finite sum formula with $n = 4$ :

S_4 = \frac{8\!\left(1 - \left(\tfrac{2}{3}\right)^4\right)}{1 - \tfrac{2}{3}} = \frac{8\!\left(1 - \tfrac{16}{81}\right)}{\tfrac{1}{3}} = 24 \times \frac{65}{81} = \frac{1560}{81} = \frac{520}{27}.

Answer: $r = \dfrac{2}{3}$ , $S_4 = \dfrac{520}{27}$ .

Worked example 2 — Recurrence relation and its limit

Problem. A sequence is defined by $u_1 = 1$ and $u_{n+1} = \dfrac{1}{2}u_n + 3$ . Describe the behaviour of the sequence, and find the exact value of the limit $L$ to which it converges.

Solution.

Generating the first few terms (GC table, or by hand):

u_1 = 1, \quad u_2 = 3.5, \quad u_3 = 4.75, \quad u_4 = 5.375, \quad u_5 = 5.6875, \; \ldots

The terms are increasing and appear to approach a value near $6$ .

For the exact limit: if $u_n \to L$ , then $u_{n+1} \to L$ as well, so taking limits on both sides of the recurrence,

L = \frac{1}{2}L + 3 \implies \frac{1}{2}L = 3 \implies L = 6.

Answer: the sequence is increasing and converges to $L = 6$ .

The same two-step pattern — observe behaviour with the GC, then solve $L = f(L)$ for the exact limit — handles almost every recurrence question in the current syllabus.

Worked example 3 — Arithmetic and geometric in the same problem

Problem. The 3rd, 6th, and 14th terms of an AP with first term $a$ and common difference $d > 0$ form a GP. Find the common ratio of the GP.

Solution.

The three AP terms are:

T_3 = a + 2d, \quad T_6 = a + 5d, \quad T_{14} = a + 13d.

For these to form a GP the ratio between consecutive terms must be equal:

\frac{a+5d}{a+2d} = \frac{a+13d}{a+5d}.

Cross-multiplying:

(a+5d)^2 = (a+2d)(a+13d).

Expanding:

a^2 + 10ad + 25d^2 = a^2 + 15ad + 26d^2.

Simplifying (subtract $a^2$ from both sides):

10ad + 25d^2 = 15ad + 26d^2 \implies -5ad - d^2 = 0 \implies d(-5a - d) = 0.

Since $d > 0$ , we have $d \neq 0$ , so $-5a - d = 0 \implies a = -\dfrac{d}{5}$ .

The common ratio is:

r = \frac{a+5d}{a+2d} = \frac{-\tfrac{d}{5}+5d}{-\tfrac{d}{5}+2d} = \frac{\tfrac{24d}{5}}{\tfrac{9d}{5}} = \frac{24}{9} = \frac{8}{3}.

Answer: common ratio $= \dfrac{8}{3}$ .

Common mistakes

Forgetting the convergence condition. Always state $|r| < 1$ when using $S_\infty$ ; omitting it costs a mark.
Sign errors in the ratio formula. $S_n = \dfrac{a(1-r^n)}{1-r}$ and $\dfrac{a(r^n - 1)}{r-1}$ are equivalent — pick one and be consistent; mixing the two forms causes sign errors.
Assuming a recurrence converges. Solving $L = f(L)$ only finds the candidate limit — the question usually expects you to show behaviour (e.g. from the GC table) before claiming convergence.
Treating an AP sum as a GP or vice versa. Read the question carefully — “common difference” signals AP, “common ratio” signals GP.