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Additional Mathematics

Sequences and Series

PDF
Matthew Williams
|May 16, 2026|6 min read
ArithmeticConvergenceGeometricPaper 01Paper 02Section 1SequencesSeries

Arithmetic and geometric sequences, nth-term formulas, sums of finite series, convergence and divergence, sum to infinity, and real-world applications including compound interest.

A sequence is an ordered list of numbers following a rule. A series is the sum of the terms of a sequence. Paper 01 allocates three items to this section; Paper 02 frequently integrates sequences into multi-part algebra questions.

Notation

The terms of a sequence are written u1,u2,u3,…,unu_1, u_2, u_3, \ldots, u_nu1​,u2​,u3​,…,un​, where unu_nun​ is the general term (also called the nnnth term). The sum of the first nnn terms is written SnS_nSn​.

Sigma notation compresses series:

Sn=∑r=1nur=u1+u2+⋯+unS_n = \sum_{r=1}^{n} u_r = u_1 + u_2 + \cdots + u_nSn​=∑r=1n​ur​=u1​+u2​+⋯+un​

The letter rrr is the index of summation. Key rules: ∑(ur+vr)=∑ur+∑vr\sum (u_r + v_r) = \sum u_r + \sum v_r∑(ur​+vr​)=∑ur​+∑vr​ and ∑c ur=c∑ur\sum c\,u_r = c\sum u_r∑cur​=c∑ur​.

Arithmetic Sequences

An arithmetic sequence has a constant difference ddd between consecutive terms.

u1,u1+d,u1+2d,…u_1,\quad u_1 + d,\quad u_1 + 2d,\quad \ldotsu1​,u1​+d,u1​+2d,…

The nnnth term is:

un=a+(n−1)du_n = a + (n-1)dun​=a+(n−1)d

where a=u1a = u_1a=u1​ is the first term and ddd is the common difference (which may be negative).

Identifying arithmetic sequences: check that u2−u1=u3−u2=du_2 - u_1 = u_3 - u_2 = du2​−u1​=u3​−u2​=d (constant).

Example

Find the 25th term of the sequence 7,11,15,19,…7, 11, 15, 19, \ldots7,11,15,19,…

a=7a = 7a=7, d=4d = 4d=4. u25=7+24(4)=7+96=103\quad u_{25} = 7 + 24(4) = 7 + 96 = 103u25​=7+24(4)=7+96=103

Example

An arithmetic sequence has u5=17u_5 = 17u5​=17 and u12=38u_{12} = 38u12​=38. Find aaa and ddd.

u12−u5=7d=38−17=21⇒d=3u_{12} - u_5 = 7d = 38 - 17 = 21 \Rightarrow d = 3u12​−u5​=7d=38−17=21⇒d=3 a=u5−4d=17−12=5a = u_5 - 4d = 17 - 12 = 5a=u5​−4d=17−12=5

Sum of an Arithmetic Series

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d]Sn​=2n​[2a+(n−1)d]

An equivalent form when the last term l=unl = u_nl=un​ is known:

Sn=n2(a+l)S_n = \frac{n}{2}(a + l)Sn​=2n​(a+l)

Example

Find the sum of the first 20 terms of 5,9,13,17,…5, 9, 13, 17, \ldots5,9,13,17,…

a=5a = 5a=5, d=4d = 4d=4, n=20n = 20n=20.

S20=202[2(5)+19(4)]=10[10+76]=10×86=860S_{20} = \frac{20}{2}[2(5) + 19(4)] = 10[10 + 76] = 10 \times 86 = 860S20​=220​[2(5)+19(4)]=10[10+76]=10×86=860

Divergence of Arithmetic Series

Every arithmetic series diverges unless d=0d = 0d=0. As nnn grows without bound, SnS_nSn​ grows without bound (positively or negatively). There is no finite sum to infinity for an arithmetic series.

Geometric Sequences

A geometric sequence has a constant ratio rrr between consecutive terms.

u1,u1r,u1r2,…u_1,\quad u_1 r,\quad u_1 r^2,\quad \ldotsu1​,u1​r,u1​r2,…

The nnnth term is:

un=arn−1u_n = ar^{n-1}un​=arn−1

Identifying geometric sequences: check that u2/u1=u3/u2=ru_2/u_1 = u_3/u_2 = ru2​/u1​=u3​/u2​=r (constant).

Example

Find the 8th term of 3,6,12,24,…3, 6, 12, 24, \ldots3,6,12,24,…

a=3a = 3a=3, r=2r = 2r=2. u8=3×27=3×128=384\quad u_8 = 3 \times 2^7 = 3 \times 128 = 384u8​=3×27=3×128=384

Sum of a Geometric Series

For r≠1r \neq 1r=1:

Sn=a(rn−1)r−1(use when r>1)S_n = \frac{a(r^n - 1)}{r - 1} \qquad \text{(use when } r > 1\text{)}Sn​=r−1a(rn−1)​(use when r>1)

Sn=a(1−rn)1−r(use when ∣r∣<1, avoids negatives)S_n = \frac{a(1 - r^n)}{1 - r} \qquad \text{(use when } |r| < 1\text{, avoids negatives)}Sn​=1−ra(1−rn)​(use when ∣r∣<1, avoids negatives)

Both forms are equivalent. Choose whichever keeps the numerator positive to reduce sign errors.

Example

Find the sum of the first 6 terms of 2,6,18,54,…2, 6, 18, 54, \ldots2,6,18,54,…

a=2a = 2a=2, r=3r = 3r=3, n=6n = 6n=6.

S6=2(36−1)3−1=2(729−1)2=728S_6 = \frac{2(3^6 - 1)}{3 - 1} = \frac{2(729 - 1)}{2} = 728S6​=3−12(36−1)​=22(729−1)​=728

Convergence and Sum to Infinity

A geometric series converges (has a finite sum to infinity) if and only if ∣r∣<1|r| < 1∣r∣<1.

When ∣r∣≥1|r| \geq 1∣r∣≥1, the terms do not approach zero and the series diverges.

For a convergent geometric series:

S∞=a1−r(∣r∣<1)S_\infty = \frac{a}{1 - r} \qquad (|r| < 1)S∞​=1−ra​(∣r∣<1)

Example

Find the sum to infinity of 12+6+3+1.5+⋯12 + 6 + 3 + 1.5 + \cdots12+6+3+1.5+⋯

a=12a = 12a=12, r=12r = \frac{1}{2}r=21​. Since ∣r∣=12<1|r| = \frac{1}{2} < 1∣r∣=21​<1, the series converges.

S∞=121−12=1212=24S_\infty = \frac{12}{1 - \frac{1}{2}} = \frac{12}{\frac{1}{2}} = 24S∞​=1−21​12​=21​12​=24

Exam Tip

Before applying S∞S_\inftyS∞​, always state explicitly that ∣r∣<1|r| < 1∣r∣<1. Applying the formula to a divergent series (∣r∣≥1|r| \geq 1∣r∣≥1) gives a meaningless result and loses method marks.

Finding the Common Ratio from a Convergence Condition

Example

The first term of a geometric series is 888 and its sum to infinity is 202020. Find the common ratio.

S∞=a1−r⇒20=81−r⇒1−r=820=25⇒r=35S_\infty = \frac{a}{1-r} \Rightarrow 20 = \frac{8}{1-r} \Rightarrow 1-r = \frac{8}{20} = \frac{2}{5} \Rightarrow r = \frac{3}{5}S∞​=1−ra​⇒20=1−r8​⇒1−r=208​=52​⇒r=53​

Check: ∣r∣=35<1|r| = \frac{3}{5} < 1∣r∣=53​<1 ✓

Comparing Arithmetic and Geometric

FeatureArithmeticGeometric
PatternAdd constant dddMultiply by constant rrr
nnnth terma+(n−1)da + (n-1)da+(n−1)darn−1ar^{n-1}arn−1
Sum to nnn termsn2[2a+(n−1)d]\frac{n}{2}[2a+(n-1)d]2n​[2a+(n−1)d]a(rn−1)r−1\frac{a(r^n-1)}{r-1}r−1a(rn−1)​
Converges?No (unless d=0d=0d=0)Yes, if ∣r∣<1\lvert r\rvert<1∣r∣<1
Sum to infinityDoes not exista1−r\frac{a}{1-r}1−ra​ when ∣r∣<1\lvert r\rvert<1∣r∣<1

Real-World Applications

Repeated percentage change produces a geometric sequence. A quantity starting at AAA and changing by a fixed factor kkk each period has value Akn−1Ak^{n-1}Akn−1 after (n−1)(n-1)(n−1) periods.

Example

A car is purchased for 20,000 and depreciates by 15% each year. Find its value after 4 years.

Each year the value is multiplied by 1−0.15=0.851 - 0.15 = 0.851−0.15=0.85.

Value after 4 years: 20,000×(0.85)4≈10,44020{,}000 \times (0.85)^4 \approx 10{,}44020,000×(0.85)4≈10,440

Example

An investment of 5,000 earns compound interest at 6% per annum. How many complete years until it doubles?

After nnn years: value =5000×(1.06)n= 5000 \times (1.06)^n=5000×(1.06)n. Set this ≥10,000\geq 10{,}000≥10,000:

(1.06)n≥2(1.06)^n \geq 2(1.06)n≥2

nlog⁡1.06≥log⁡2⇒n≥log⁡2log⁡1.06≈11.9n \log 1.06 \geq \log 2 \Rightarrow n \geq \frac{\log 2}{\log 1.06} \approx 11.9nlog1.06≥log2⇒n≥log1.06log2​≈11.9

The investment first exceeds double after 12 complete years.

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Surds, Indices, and Logarithms
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Coordinate Geometry