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Intermediate25 min11 / 68

Geometric sequences

Geometric sequences for CSCA: the general term, Sₙ for q ≠ 1 and q = 1, properties of products and block sums, the sum of an infinite decreasing geometric series, finding aₙ from Sₙ and mixed arithmetic–geometric problems.

Check yourself
In this lesson you will learn
  • Find any term, q and a₁ of a geometric sequence and decide its sign pattern and monotonicity
  • Compute Sₙ with the right formula (q ≠ 1 or q = 1) and use the product and block-sum properties
  • Find the sum of an infinite decreasing geometric series, including repeating decimals
  • Find aₙ from Sₙ and solve mixed arithmetic–geometric problems, including sums by shifted subtraction

Drop a ball from 10 m. Each time it bounces back to half of its previous height: 5 m, 2.5 m, 1.25 m, … In theory it bounces infinitely many times, yet the total distance it travels is finite: exactly 30 m. The heights form a geometric sequence, in which each term is the previous one times the same number. In the lesson «Arithmetic sequences» you added a fixed number each time; here you multiply, so the terms grow or shrink exponentially, and a sum can even have infinitely many terms. This lesson covers everything CSCA asks about geometric sequences: the general term, the sum, the properties, infinite sums, finding aₙ from Sₙ and mixed problems.

Definition and the general term

Definition
Geometric sequence (等比数列)

A sequence in which every term divided by the previous one gives the same number q: aₙ₊₁/aₙ = q for all n. The number q is the common ratio (公比; many English books write r). No term and no ratio can be 0: a₁ ≠ 0 and q ≠ 0.

aₙ = a₁qⁿ⁻¹ aₙ = aₘqⁿ⁻ᵐ qⁿ⁻ᵐ = aₙ/aₘaₙ = a₁qⁿ⁻¹ aₙ = aₘqⁿ⁻ᵐ qⁿ⁻ᵐ = aₙ/aₘ
where:
  • a₁the first term (首项)
  • qthe common ratio, q ≠ 0
  • n, mpositions of terms

From a₁ to aₙ you multiply by q exactly n − 1 times. Two known terms give qⁿ⁻ᵐ; when n − m is even, q is found only up to its sign.

Example 1: the general term in three quick cases

1) a₁ = 3 and q = −2. Find a₇.
2) a₂ = 12 and a₅ = −96. Find q and a₁.
3) a₃ = 2 and a₇ = 18. Find a₅.

Show solution
1) a₇ = 3 · (−2)⁶ = 3 · 64 = 192.
2) q³ = a₅/a₂ = −96/12 = −8, so q = −2 and a₁ = a₂/q = −6. Check: a₅ = −6 · (−2)⁴ = −96.
3) q⁴ = 18/2 = 9, so q² = 3 (q² cannot be −3). a₅ = a₃q² = 2 · 3 = 6. The value −6 is impossible: a₃ and a₅ differ by the factor q² > 0, so they have the same sign.

Sign and monotonicity. Since aₙ = a₁qⁿ⁻¹, the sequence behaves like an exponential function of n (see «Power and exponential functions»). If q > 1 the terms move away from 0, if 0 < q < 1 they approach 0, if q = 1 the sequence is constant, and if q < 0 the signs alternate and the sequence is not monotonic. Terms whose positions have the same parity (both odd or both even) always have the same sign.

qa₁ > 0a₁ < 0
q > 1increasingdecreasing
0 < q < 1decreasing, terms → 0increasing, terms → 0
q = 1constantconstant
q < 0signs alternatesigns alternate
How a geometric sequence behaves depends on q and on the sign of a₁.
Definition
Geometric mean (等比中项)

If a, G, b are consecutive terms of a geometric sequence, then G² = ab, so G = ±√(ab). A geometric mean exists only when ab > 0, and both signs are possible unless other terms fix the sign. Compare: the arithmetic mean of a and b is always the single number (a + b)/2.

Example 2 (CSCA style): the sign of a middle term

The numbers −2, a, b, c, −32 form a geometric sequence. What is b?
A) 8 B) −8 C) ±8 D) −17

Show solution
b is the geometric mean of −2 and −32: b² = (−2)(−32) = 64, so b = ±8. But b = −2 · q² with q² > 0, so b has the sign of −2: b = −8, answer B.
C (±8) stops at b² = 64; A takes the positive root; D (−17) is the arithmetic mean (−2 − 32)/2.

The sum of the first n terms and the properties

Write Sₙ and qSₙ one under the other, shifted by one place:
Sₙ = a₁ + a₁q + a₁q² + … + a₁qⁿ⁻¹
qSₙ = a₁q + a₁q² + … + a₁qⁿ⁻¹ + a₁qⁿ.
Subtracting, everything in the middle cancels: (1 − q)Sₙ = a₁ − a₁qⁿ. Dividing by 1 − q is allowed only when q ≠ 1. When q = 1 all terms equal a₁, and the sum is simply na₁. This "multiply by q and subtract" idea (错位相减) returns at the end of the lesson.

Sₙ = a₁(1 − qⁿ)/(1 − q) = (a₁ − aₙq)/(1 − q) (q ≠ 1) Sₙ = na₁ (q = 1)Sₙ = a₁(1 − qⁿ)/(1 − q) = (a₁ − aₙq)/(1 − q) (q ≠ 1) Sₙ = na₁ (q = 1)
where:
  • Sₙthe sum of the first n terms (前n项和)
  • aₙthe last term added
  • qthe common ratio

The sum of a geometric sequence. For q > 1 the same formula written as a₁(qⁿ − 1)/(q − 1) avoids minus signs. When q is unknown, always ask whether q = 1 is possible.

Example 3: sums in three quick cases

1) 1 + 2 + 4 + … + 2⁹.
2) 81 + 27 + 9 + 3 + 1.
3) 4 − 12 + 36 − … + 324.

Show solution
1) a₁ = 1, q = 2, n = 10: S₁₀ = (2¹⁰ − 1)/(2 − 1) = 1023.
2) a₁ = 81, q = 1/3, n = 5: S₅ = 81 · (1 − 1/243)/(1 − 1/3) = 81 · (242/243) · (3/2) = 121. Adding directly gives the same.
3) a₁ = 4, q = −3, last term aₙ = 324: Sₙ = (a₁ − aₙq)/(1 − q) = (4 + 972)/4 = 244. The form with aₙ saves finding n (here n = 5).
Example 4 (CSCA style): do not forget q = 1

A geometric sequence has a₃ = 4 and S₃ = 12. What is q?
A) 1 B) −1/2 C) 1 or −1/2 D) 1 or −2

Show solution
a₁ = 4/q² and a₂ = 4/q, so 4/q² + 4/q + 4 = 12, i.e. 1/q² + 1/q − 2 = 0. With t = 1/q: t² + t − 2 = 0, t = 1 or t = −2, so q = 1 or q = −1/2, answer C.
Check: q = 1 gives 4 + 4 + 4 = 12; q = −1/2 gives 16 − 8 + 4 = 12. B appears if you use a₁(1 − q³)/(1 − q), which silently excludes q = 1; A misses the second root; D forgets to turn t back into q = 1/t.
m + n = p + t ⇒ aₘaₙ = aₚaₜ aₙ² = aₙ₋ₖaₙ₊ₖ (S₂ₖ − Sₖ) : Sₖ = qᵏ
where:
  • m, n, p, tpositions with equal sums
  • kthe number of steps to the left and right of aₙ; the length of a block
  • qᵏthe ratio of neighbouring blocks of k terms

The product version of the index property of arithmetic sequences. Consecutive blocks of k terms, Sₖ, S₂ₖ − Sₖ, S₃ₖ − S₂ₖ, form a geometric sequence with ratio qᵏ (when Sₖ ≠ 0). Also a₁a₂…a₂ₖ₋₁ = aₖ²ᵏ⁻¹: the product of an odd number of terms is the middle term to that power.

Example 5: properties in two quick cases

1) A geometric sequence has positive terms and a₂a₁₀ = 9. Find a₆ and a₄a₈.
2) S₂ = 3 and S₄ = 15. Find S₆.

Show solution
1) 2 + 10 = 6 + 6, so a₆² = 9 and, the terms being positive, a₆ = 3; also 4 + 8 = 2 + 10, so a₄a₈ = 9.
2) The blocks S₂ = 3, S₄ − S₂ = 12, S₆ − S₄ form a geometric sequence with ratio q² = 4, so S₆ − S₄ = 48 and S₆ = 63. (Both q = 2, a₁ = 1 and q = −2, a₁ = −3 fit the data, and both give 63.)
Example 6 (CSCA style): pairs of terms

In a geometric sequence a₁ + a₂ = 3 and a₃ + a₄ = 12. What is a₇ + a₈?
A) 48 B) 96 C) 384 D) 192

Show solution
a₃ + a₄ = q²(a₁ + a₂), so q² = 12/3 = 4. Each pair is 4 times the previous one: a₅ + a₆ = 48, a₇ + a₈ = 192. In one step: a₇ + a₈ = q⁶(a₁ + a₂) = 64 · 3 = 192, answer D.
A (48) is a₅ + a₆, one step too early; C (384) uses q⁷ = 128 instead of q⁶; B (96) multiplies by 2 instead of q² = 4 in the last step.

The infinite decreasing geometric series

Definition
Sum of an infinite decreasing geometric sequence (无穷递缩等比数列)

If |q| < 1 and q ≠ 0, then qⁿ → 0 as n grows, so Sₙ = a₁(1 − qⁿ)/(1 − q) gets as close as we like to a₁/(1 − q). This limit is called the sum of all terms of the infinite geometric sequence. For |q| ≥ 1 the terms do not shrink to 0, and there is no finite sum.

S = a₁/(1 − q) (0 < |q| < 1)S = a₁/(1 − q) (0 < |q| < 1)
where:
  • Sthe sum of all terms, the limit of Sₙ
  • a₁the first term
  • qthe common ratio, |q| < 1

The sum of an infinite decreasing geometric series. With q < 0 the terms alternate in sign, and the formula still works. Sanity check: an infinite sum of positive terms is always larger than a₁.

Interactive
Loading simulation…
The curve joins the partial sums Sₙ (read it only at whole n), and the horizontal line is S = a₁/(1 − q). Move q to 0.8: the sums approach the line more slowly and S grows (a₁ = 5, q = 0.8 gives S = 25); move q to 0.1 and S₂ is already almost S. Negative q is not shown here, because qˣ is undefined for fractional x.
Example 7: infinite sums in three quick cases

1) 8 − 4 + 2 − 1 + …
2) Write 0.363636… (the block 36 repeats) as a fraction.
3) The ball from the start: dropped from 10 m, each bounce reaches half of the previous height. Find the total distance.

Show solution
1) a₁ = 8, q = −1/2: S = 8/(1 + 1/2) = 16/3.
2) 0.3636… = 36/100 + 36/100² + … with a₁ = 36/100 and q = 1/100: S = (36/100)/(99/100) = 36/99 = 4/11.
3) First 10 m down, then every bounce is travelled twice (up and down): 5 + 5, 2.5 + 2.5, … The distance is 10 + 2 · 5/(1 − 1/2) = 10 + 20 = 30 m.
Example 8 (CSCA style): the sum of the squares

An infinite geometric sequence has first term 4, and the sum of all its terms is 6. What is the sum of the squares of all its terms?
A) 36 B) 24 C) 18 D) 12

Show solution
4/(1 − q) = 6 ⇒ 1 − q = 2/3 ⇒ q = 1/3. The squares 16, 16/9, 16/81, … form a geometric sequence with first term 16 and ratio q² = 1/9, so their sum is 16/(1 − 1/9) = 16 · 9/8 = 18, answer C.
A (36) squares the sum 6; B (24) keeps the ratio 1/3 for the squares, 16/(1 − 1/3); D (12) uses 1 + q instead of 1 − q, 16/(1 + 1/3).

Finding aₙ from Sₙ, and mixed problems

aₙ = S₁ (n = 1); aₙ = Sₙ − Sₙ₋₁ (n ≥ 2)
where:
  • Sₙthe sum of the first n terms, given as a formula
  • Sₙ₋₁the same formula with n − 1 in place of n

This works for every sequence, not only geometric ones. Sₙ − Sₙ₋₁ is the last term aₙ, but only for n ≥ 2, because S₀ is not part of the rule. Always check whether the formula for n ≥ 2 also gives a₁ = S₁; if not, the answer is piecewise.

  1. 1
    a₁ first

    Compute a₁ = S₁.

  2. 2
    Subtract

    For n ≥ 2 write aₙ = Sₙ − Sₙ₋₁ and simplify.

  3. 3
    Test n = 1

    Put n = 1 into the result. If it gives S₁, one formula serves all n; otherwise write aₙ piecewise.

  4. 4
    Relations such as Sₙ = 2aₙ − 1

    Write the same relation for n − 1 and subtract: Sₙ − Sₙ₋₁ = aₙ turns it into a rule linking aₙ and aₙ₋₁.

Example 9: aₙ from Sₙ

Find aₙ: 1) Sₙ = 2 · 3ⁿ − 2; 2) Sₙ = 2ⁿ + 1; 3) Sₙ = n² + 3n.

Show solution
1) a₁ = S₁ = 4. For n ≥ 2: aₙ = 2 · 3ⁿ − 2 · 3ⁿ⁻¹ = 4 · 3ⁿ⁻¹, and n = 1 gives 4 as well: aₙ = 4 · 3ⁿ⁻¹, geometric with q = 3 (the shape A − A · qⁿ with A = −2).
2) a₁ = S₁ = 3. For n ≥ 2: aₙ = 2ⁿ − 2ⁿ⁻¹ = 2ⁿ⁻¹, but n = 1 would give 1 ≠ 3. So a₁ = 3 and aₙ = 2ⁿ⁻¹ for n ≥ 2: not a geometric sequence (the constant +1 is not −1).
3) a₁ = 4; for n ≥ 2: aₙ = n² + 3n − (n − 1)² − 3(n − 1) = 2n + 2, which also gives a₁ = 4: aₙ = 2n + 2, arithmetic with d = 2 (the shape An² + Bn from «Arithmetic sequences»).
Example 10 (CSCA style): Sₙ = 3aₙ − 3

The sum of the first n terms of {aₙ} satisfies Sₙ = 3aₙ − 3 for every n. What is aₙ?
A) (3/2)ⁿ B) (3/2)ⁿ⁻¹ C) 3ⁿ⁻¹ D) (2/3)ⁿ

Show solution
n = 1: a₁ = 3a₁ − 3 ⇒ a₁ = 3/2. For n ≥ 2 subtract Sₙ₋₁ = 3aₙ₋₁ − 3: aₙ = 3aₙ − 3aₙ₋₁ ⇒ 2aₙ = 3aₙ₋₁ ⇒ aₙ/aₙ₋₁ = 3/2. So {aₙ} is geometric with a₁ = q = 3/2: aₙ = (3/2)(3/2)ⁿ⁻¹ = (3/2)ⁿ, answer A.
Shortcut: only A gives a₁ = 3/2 at n = 1. B has the right ratio but first term 1; C takes q = 3 from the coefficient; D turns the ratio upside down.

Mixed problems combine the two kinds of sequences. Typical links: three numbers that are arithmetic become geometric after a change (use A = (a + c)/2 and G² = ac); some terms of an arithmetic sequence form a geometric sequence; a recursion aₙ₊₁ = paₙ + r becomes geometric after adding the constant λ = r/(p − 1): aₙ₊₁ + λ = p(aₙ + λ); and sums of (arithmetic) × (geometric) terms are found by shifted subtraction.

Example 11: three mixed problems

1) Three numbers form an arithmetic sequence with sum 9. If 1, 1 and 3 are added to them respectively, the results form a geometric sequence. Find the numbers.
2) An arithmetic sequence with d ≠ 0 has a₂, a₄, a₈ forming a geometric sequence. Find its common ratio.
3) a₁ = 2 and aₙ₊₁ = 3aₙ − 2. Find aₙ.

Show solution
1) Write the numbers as 3 − d, 3, 3 + d. Then (4 − d), 4, (6 + d) is geometric: (4 − d)(6 + d) = 16 ⇒ d² + 2d − 8 = 0 ⇒ d = 2 or d = −4. 1, 3, 5 (giving 2, 4, 8) or 7, 3, −1 (giving 8, 4, 2).
2) (a₁ + 3d)² = (a₁ + d)(a₁ + 7d) ⇒ 6a₁d + 9d² = 8a₁d + 7d² ⇒ 2d² = 2a₁d ⇒ d = a₁ (d ≠ 0). Then a₂ = 2a₁, a₄ = 4a₁, a₈ = 8a₁: ratio 2.
3) aₙ₊₁ − 1 = 3(aₙ − 1), so {aₙ − 1} is geometric with first term 1 and ratio 3: aₙ − 1 = 3ⁿ⁻¹, aₙ = 3ⁿ⁻¹ + 1. Check: a₂ = 3 · 2 − 2 = 4 = 3 + 1.
  1. 1
    Write Tₙ

    Shifted subtraction (错位相减), for example Tₙ = 1 · 2 + 2 · 2² + 3 · 2³ + … + n · 2ⁿ.

  2. 2
    Multiply by q

    2Tₙ = 1 · 2² + 2 · 2³ + … + (n − 1) · 2ⁿ + n · 2ⁿ⁺¹, written one place to the right.

  3. 3
    Subtract

    Tₙ − 2Tₙ = 2 + 2² + … + 2ⁿ − n · 2ⁿ⁺¹: the middle becomes a plain geometric sum, and the last term comes with a minus sign.

  4. 4
    Finish

    −Tₙ = (2ⁿ⁺¹ − 2) − n · 2ⁿ⁺¹, so Tₙ = (n − 1) · 2ⁿ⁺¹ + 2. Check: n = 1 gives 0 + 2 = 2 = T₁; n = 2 gives 8 + 2 = 10 = 2 + 8.

How CSCA asks about this

  • Two facts → the sequence: two terms give qⁿ⁻ᵐ (watch the ± when n − m is even); a term and a sum often lead to a quadratic in q or 1/q.
  • Products and blocks: aₘaₙ = aₚaₜ, sums of logarithms (log a₁ + … + log aₙ = the log of a product), block sums with ratio qᵏ.
  • q = 1 and signs: options like "1 or −1/2" test whether you remembered q = 1; the geometric mean ±√(ab) versus the sign fixed by q².
  • Infinite sums: S = a₁/(1 − q), repeating decimals, bouncing balls, sums of squares (ratio q²).
  • aₙ from Sₙ: Sₙ = A − A · qⁿ, Sₙ = kaₙ + c; check n = 1.
  • Mixed: arithmetic and geometric conditions together, aₙ₊₁ = paₙ + r, shifted subtraction.
  • Time-savers (75 s per item): test the options with n = 1 and n = 2; in "find q" items substitute the options; sanity check: an infinite sum of positive terms is larger than a₁.
Term中文Pinyin
geometric sequence等比数列děngbǐ shùliè
common ratio公比gōngbǐ
first term首项shǒuxiàng
geometric mean等比中项děngbǐ zhōngxiàng
sum of the first n terms前n项和qián n xiàng hé
constant sequence常数列chángshùliè
alternating (swinging) sequence摆动数列bǎidòng shùliè
infinite decreasing geometric sequence无穷递缩等比数列wúqióng dìsuō děngbǐ shùliè
limit极限jíxiàn
shifted subtraction (multiply by q and subtract)错位相减法cuòwèi xiāngjiǎn fǎ
recursive formula递推公式dìtuī gōngshì
exponential growth指数增长zhǐshù zēngzhǎng
Key terms in English and Chinese: you may take the test in either language.

Key points

  • aₙ = a₁qⁿ⁻¹ = aₘqⁿ⁻ᵐ with a₁ ≠ 0 and q ≠ 0; terms at positions of the same parity share a sign.
  • Sₙ = a₁(1 − qⁿ)/(1 − q) = (a₁ − aₙq)/(1 − q) for q ≠ 1 and Sₙ = na₁ for q = 1.
  • m + n = p + t ⇒ aₘaₙ = aₚaₜ; the geometric mean satisfies G² = ab; block sums have ratio qᵏ.
  • For 0 < |q| < 1 the infinite sum is S = a₁/(1 − q).
  • a₁ = S₁ and aₙ = Sₙ − Sₙ₋₁ (n ≥ 2); a geometric sum has the shape Sₙ = A − A · qⁿ.
  • Arithmetic × geometric sums: multiply by q, shift, subtract.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
Which of these sequences is geometric?

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