IB Maths AI HLSequences & SeriesPaper 1 & 2common ratio r~7 min read
Geometric Sequences & Series
A geometric sequence multiplies by the same fixed number each time — the common ratio. As well as formulae for any term and for the sum of n terms, geometric series add a new idea: when the ratio is small enough, even an infinite sum has a finite total.
📘 What you need to know
A geometric sequence multiplies by a constant — the common ratio r.
It is increasing if r > 1, decreasing if 0 < r < 1, and alternating in sign if r < 0.
nth term: un = u1rn−1.
Sum of the first n terms: Sn = u1(rn − 1)r − 1.
Sum to infinity: if |r| < 1 the series converges to S∞ = u11 − r.
Find r by dividing a term by the previous one; find n with logarithms.
Geometric sequences
A geometric sequence is one where each term is the previous term multiplied by a fixed number, the common ratio r. For 2, 6, 18, 54, … the ratio is r = 3. You find r by dividing any term by the one before it.
The terms 6, 12, 24, 48 each double the one before — a constant ratio r = 2 makes the sequence geometric.
nth term of a geometric sequenceun = u1rn−1u1 = first term, r = common ratio — given in the formula booklet
The sum of a geometric series
A geometric series is the sum of the terms of a geometric sequence. There are two equivalent forms for Sn.
Sum of the first n termsSn = u1(rn − 1)r − 1 = u1(1 − rn)1 − rthe first form is tidier when r > 1, the second when r < 1 — both in the booklet
If the number of terms n is unknown, substituting into the formula leaves rn, so you solve for n with logarithms.
The sum to infinity
If you keep adding terms forever, the total either grows without bound or settles on a finite value. The deciding factor is the size of r.
Convergence: when |r| < 1 each term shrinks, the partial sums approach a limit, and the series converges. When |r| ≥ 1 it diverges — there is no finite sum to infinity.
Sum to infinity (only when |r| < 1)S∞ = u11 − rvalid only for |r| < 1 — the formula and condition are both in the booklet
🧠Recipe — geometric sequences and series
Identify u1 and r — find r by dividing any term by the previous one.
For a term, use un = u1rn−1.
For a sum of n terms, use Sn = u1(rn−1)r−1 — or the (1−rn) form when r < 1.
If n is unknown, substitute into the formula and use logarithms to solve for n.
For an infinite sum, check |r| < 1 first, then use S∞ = u11−r.
Worked examples
WE 1
Finding a term
A geometric sequence has first term 3 and common ratio 2. Find the 7th term.
use un = u1rn−1u7 = 3 × 27−1 = 3 × 26= 3 × 64u7 = 192the power is n − 1 — the first term has been multiplied by r six times.
WE 2
Finding r and u₁ from two terms
The 2nd term of a geometric sequence is 12 and the 5th term is 96. Find the common ratio and the first term.
write each term, then divide to remove u1u2 = u1r = 12, u5 = u1r4 = 96u5 ÷ u2: r3 = 96 ÷ 12 = 8cube root, then back-substituter = 2; u1 = 12 ÷ 2 = 6r = 2, u1 = 6dividing the two terms cancels u1 and leaves a power of r.
WE 3
Summing a geometric series
Find the sum of the first 8 terms of the geometric sequence 5, 10, 20, 40, …
u1 = 5, r = 2, n = 8S8 = 5(28 − 1)2 − 1= 5(256 − 1)1 = 5 × 255S8 = 1275r > 1, so the (rn − 1) form keeps everything positive.
WE 4
Finding n with logarithms
A geometric sequence has first term 3 and common ratio 1.5. Find the smallest value of n for which un exceeds 100.
set up the inequality3 × 1.5n−1 > 1001.5n−1 > 33.33…take logs to free nn − 1 > log1.533.33… = 8.648…n > 9.648…smallest n = 10n must be a whole number, so round the inequality up to the next integer.
WE 5
Sum to infinity
The first three terms of a geometric sequence are 20, 15, 11.25. (a) Show that the series converges. (b) Find the sum to infinity.
(a) find r by dividing termsr = 15 ÷ 20 = 0.75|0.75| < 1 ⇒ the series converges(b) use S∞ = u1 ÷ (1 − r)S∞ = 201 − 0.75 = 200.25(a) converges · (b) S∞ = 80always state the |r| < 1 check before using the sum-to-infinity formula.
WE 6
Full question: term, finite sum and infinite sum
A geometric sequence has first term 64 and common ratio 0.5. (a) Find the 6th term. (b) Find the sum of the first 6 terms. (c) Find the sum to infinity.
(a) u6 = 64 × 0.55= 64 × 0.03125 = 2(b) r < 1, use the (1 − rn) formS6 = 64(1 − 0.56)1 − 0.5 = 630.5 = 126(c) |r| < 1, so S∞ = 64 ÷ (1 − 0.5)(a) 2 · (b) 126 · (c) S∞ = 128S6 = 126 is already close to S∞ = 128 — the partial sums home in on the limit.
💡 Top tips
Find r by dividing a term by the one before — never by subtracting.
The nth term uses rn−1, not rn.
Pick the sum form to suit the ratio: (rn−1) for r > 1, (1−rn) for r < 1.
An unknown n sits inside a power — logarithms are how you get it out.
Only use S∞ after checking |r| < 1 — otherwise the series has no finite sum.
âš Common mistakes
Subtracting terms to find r — the common ratio comes from dividing.
Using rn in the nth-term formula instead of rn−1.
Applying S∞ when |r| ≥ 1 — that series diverges, with no finite total.
Forgetting a negative r when only a power of r is known — check whether the sequence alternates.
Rounding a logarithm answer the wrong way — for “exceeds”, round the value of n up.
Next up: Applications of Sequences & Series — spotting which model a worded problem needs. A fixed amount added each time is arithmetic; a fixed factor multiplied each time is geometric, as with compound interest, growth and decay.
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