Mathematics · Sequences
Geometric Sequence & Series
Self-Study Notes · 20 Problems · Basic → Advanced
⚡ Super Quick Memory Points
RATIO
Common ratio r = next ÷ prev
Always the same!
Always the same!
nth TERM
\(a_n = a_1 \cdot r^{n-1}\)
First × ratio^(n−1)
First × ratio^(n−1)
SUM (finite)
\(S_n = a_1 \dfrac{1-r^n}{1-r}\) when \(r \neq 1\)
SUM (infinite)
\(S_\infty = \dfrac{a_1}{1-r}\) only when \(|r| < 1\)
📐 Core Formulas — Write These Down!
| Common Ratio | \(r = \dfrac{a_{n+1}}{a_n}\) | Divide any term by the one before it |
| General Term | \(a_n = a_1 \cdot r^{n-1}\) | n-th term formula |
| Partial Sum | \(S_n = \dfrac{a_1(1-r^n)}{1-r}\) | Sum of first n terms (\(r \neq 1\)) |
| Infinite Sum | \(S = \dfrac{a_1}{1-r},\ |r|<1\) | Converges only if |r| < 1 |
✏️ ~ ~ ~ ~ ~ ~ ~ ~ ✏️
🌱 Level 1 — Basics (Q1–Q5)
⭐ Easy
1
Finding the Common Ratio
The common ratio is r = 3. Always check 2–3 pairs to be sure!
📎 Example: In 2, 6, 18, 54, … → r = 6÷2 = 3. Each term = previous × 3
What is the common ratio of the geometric sequence:
\[4,\ 12,\ 36,\ 108,\ \ldots\]
\[4,\ 12,\ 36,\ 108,\ \ldots\]
💡 HINT: r = any term ÷ previous term
📝 Explanation
Divide any consecutive terms: \(\dfrac{12}{4} = 3\), \(\dfrac{36}{12} = 3\), \(\dfrac{108}{36} = 3\).The common ratio is r = 3. Always check 2–3 pairs to be sure!
⭐ Easy
2
Identify the Geometric Sequence
A = triangular numbers, B = arithmetic (add 2), D = Fibonacci.
Which of the following is a geometric sequence?
📝 Explanation
C is geometric: \(\frac{10}{5}=\frac{20}{10}=\frac{40}{20}=2\). Constant ratio = 2 ✅A = triangular numbers, B = arithmetic (add 2), D = Fibonacci.
⭐ Easy
3
Find the Next Term
Next term = \(3 \times \dfrac{1}{3} = \mathbf{1}\) ✅
Remember: r can be a fraction! That makes the sequence shrink.
📎 Example: 3, 6, 12, ____ → next = 12 × 2 = 24
Find the next term of the sequence:
\[81,\ 27,\ 9,\ 3,\ \underline{\quad}\]
\[81,\ 27,\ 9,\ 3,\ \underline{\quad}\]
💡 KEY WORD: Decreasing ratio — r is a fraction!
📝 Explanation
Common ratio: \(r = \dfrac{27}{81} = \dfrac{1}{3}\)Next term = \(3 \times \dfrac{1}{3} = \mathbf{1}\) ✅
Remember: r can be a fraction! That makes the sequence shrink.
⭐ Easy
4
nth Term Formula
⚠️ Common mistake: using \(r^n\) instead of \(r^{n-1}\). The exponent is always n MINUS 1!
A geometric sequence has \(a_1 = 2\) and \(r = 5\).
Find \(a_4\).
Find \(a_4\).
💡 FORMULA: \(a_n = a_1 \cdot r^{n-1}\)
📝 Explanation
\(a_4 = 2 \times 5^{4-1} = 2 \times 5^3 = 2 \times 125 = \mathbf{250}\) ✅⚠️ Common mistake: using \(r^n\) instead of \(r^{n-1}\). The exponent is always n MINUS 1!
⭐ Easy
5
Negative Common Ratio
\(a_5 = 3 \times (-2)^{4} = 3 \times 16 = \mathbf{48}\) ✅
\((-2)^4 = +16\) because even power → positive sign!
Find the 5th term of the sequence:
\[3,\ -6,\ 12,\ -24,\ \underline{\quad}\]
\[3,\ -6,\ 12,\ -24,\ \underline{\quad}\]
💡 KEY: negative r → terms alternate signs (+ − + − …)
📝 Explanation
\(r = \dfrac{-6}{3} = -2\)\(a_5 = 3 \times (-2)^{4} = 3 \times 16 = \mathbf{48}\) ✅
\((-2)^4 = +16\) because even power → positive sign!
✦ ✦ ✦ ✦ ✦
📘 Level 2 — Intermediate (Q6–Q12)
⭐⭐ Medium
6
Find a₁ Given aₙ and r
\(54 = a_1 \cdot 3^3 = a_1 \cdot 27\)
\(a_1 = \dfrac{54}{27} = \mathbf{2}\) ✅
📎 Tip: Work BACKWARDS. \(a_n = a_1 \cdot r^{n-1}\) → solve for \(a_1\)
The 4th term of a geometric sequence is \(54\) and the common ratio is \(3\).
Find the first term \(a_1\).
Find the first term \(a_1\).
📝 Explanation
\(a_4 = a_1 \cdot r^3\)\(54 = a_1 \cdot 3^3 = a_1 \cdot 27\)
\(a_1 = \dfrac{54}{27} = \mathbf{2}\) ✅
⭐⭐ Medium
7
Sum of First n Terms
Or just add: \(1+2+4+8+16 = 31\)
Find \(S_5\) for the geometric sequence with \(a_1 = 1\) and \(r = 2\).
\[S_n = \frac{a_1(1 - r^n)}{1 - r}\]
\[S_n = \frac{a_1(1 - r^n)}{1 - r}\]
💡 Sequence: 1, 2, 4, 8, 16, … → just add them!
📝 Explanation
\(S_5 = \dfrac{1(1 - 2^5)}{1-2} = \dfrac{1-32}{-1} = \dfrac{-31}{-1} = \mathbf{31}\) ✅Or just add: \(1+2+4+8+16 = 31\)
⭐⭐ Medium
8
Geometric Mean
Check: \(4, 12, 36\) → \(r = 3\) ✔
📎 Geometric mean of a and b = \(\sqrt{ab}\). In a geo sequence: middle² = (left)(right)
Three numbers form a geometric sequence.
If the first is \(4\) and the third is \(36\), what is the middle term (geometric mean)?
If the first is \(4\) and the third is \(36\), what is the middle term (geometric mean)?
📝 Explanation
Geometric mean \(= \sqrt{4 \times 36} = \sqrt{144} = \mathbf{12}\) ✅Check: \(4, 12, 36\) → \(r = 3\) ✔
⭐⭐ Medium
9
Find Which Term Equals X
\(2^{n-1} = 128 = 2^7\)
\(n-1 = 7 \Rightarrow n = \mathbf{8}\) ✅
Sequence: 3, 6, 12, 24, 48, 96, 192, 384 ← 8th!
In the geometric sequence \(3,\ 6,\ 12,\ 24,\ \ldots\), which term equals \(384\)?
💡 Set \(a_n = 384\), solve: \(3 \cdot 2^{n-1} = 384\)
📝 Explanation
\(3 \cdot 2^{n-1} = 384\)\(2^{n-1} = 128 = 2^7\)
\(n-1 = 7 \Rightarrow n = \mathbf{8}\) ✅
Sequence: 3, 6, 12, 24, 48, 96, 192, 384 ← 8th!
⭐⭐ Medium
10
Infinite Geometric Series — Does it Converge?
\(|r| = \dfrac{1}{2} < 1\) → converges ✅
\(S = \dfrac{8}{1 - \frac{1}{2}} = \dfrac{8}{\frac{1}{2}} = \mathbf{16}\)
📎 Key Rule: Infinite sum exists ONLY when \(|r| < 1\). Formula: \(S = \dfrac{a_1}{1-r}\)
Find the sum of the infinite series:
\[8 + 4 + 2 + 1 + \frac{1}{2} + \cdots\]
\[8 + 4 + 2 + 1 + \frac{1}{2} + \cdots\]
📝 Explanation
\(a_1 = 8,\quad r = \dfrac{1}{2}\)\(|r| = \dfrac{1}{2} < 1\) → converges ✅
\(S = \dfrac{8}{1 - \frac{1}{2}} = \dfrac{8}{\frac{1}{2}} = \mathbf{16}\)
⭐⭐ Medium
11
Find r Given Two Terms
\(\dfrac{162}{6} = 27 = r^3\)
\(r = \sqrt[3]{27} = \mathbf{3}\) ✅
In a geometric sequence, \(a_2 = 6\) and \(a_5 = 162\).
Find the common ratio \(r\).
Find the common ratio \(r\).
💡 TRICK: \(\dfrac{a_5}{a_2} = r^{5-2} = r^3\)
📝 Explanation
\(\dfrac{a_5}{a_2} = r^{5-2} = r^3\)\(\dfrac{162}{6} = 27 = r^3\)
\(r = \sqrt[3]{27} = \mathbf{3}\) ✅
⭐⭐ Medium
12
Repeating Decimal as Infinite Series
\(S = \dfrac{\frac{6}{10}}{1 - \frac{1}{10}} = \dfrac{\frac{6}{10}}{\frac{9}{10}} = \dfrac{6}{9} = \mathbf{\dfrac{2}{3}}\) ✅
📎 Classic exam trick! \(0.\overline{3} = 0.333… = \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots\)
Express \(0.\overline{6} = 0.6666\ldots\) as a fraction using the infinite geometric series formula.
📝 Explanation
\(a_1 = \dfrac{6}{10},\quad r = \dfrac{1}{10}\)\(S = \dfrac{\frac{6}{10}}{1 - \frac{1}{10}} = \dfrac{\frac{6}{10}}{\frac{9}{10}} = \dfrac{6}{9} = \mathbf{\dfrac{2}{3}}\) ✅
★ ★ ★ ★ ★ ★ ★
🔥 Level 3 — Advanced (Q13–Q20)
⭐⭐⭐ Hard
13
Sigma Notation Sum
Or: \(3+6+12+24+48+96 = 189\)
Evaluate: \[\sum_{k=1}^{6} 3 \cdot 2^{k-1}\]
💡 This is \(S_6\) of a geometric series with \(a_1=3,\ r=2\)
📝 Explanation
\(S_6 = \dfrac{3(1-2^6)}{1-2} = \dfrac{3(1-64)}{-1} = \dfrac{3 \times (-63)}{-1} = \mathbf{189}\) ✅Or: \(3+6+12+24+48+96 = 189\)
⭐⭐⭐ Hard
14
Three Terms in Geometric Sequence
\(x^2+8x+16 = x^2+12x\)
\(16 = 4x\)
\(x = \mathbf{4}\) ✅
Check: \(4, 8, 16\) → \(r = 2\) ✔
📎 Setup: if \(x-1,\ x+1,\ x+5\) are geometric, then \((x+1)^2 = (x-1)(x+5)\)
If \(x,\ x+4,\ x+12\) are three consecutive terms of a geometric sequence, find \(x\).
💡 KEY: Middle² = First × Third
📝 Explanation
\((x+4)^2 = x(x+12)\)\(x^2+8x+16 = x^2+12x\)
\(16 = 4x\)
\(x = \mathbf{4}\) ✅
Check: \(4, 8, 16\) → \(r = 2\) ✔
⭐⭐⭐ Hard
15
Sum of Infinite Series — Find a₁
\(\dfrac{a_1}{1 - \frac{3}{4}} = 20\)
\(\dfrac{a_1}{\frac{1}{4}} = 20\)
\(a_1 = 20 \times \dfrac{1}{4} = \mathbf{5}\) ✅
The sum of an infinite geometric series is \(20\), and the common ratio is \(r = \dfrac{3}{4}\).
Find the first term \(a_1\).
Find the first term \(a_1\).
📝 Explanation
\(S = \dfrac{a_1}{1-r} = 20\)\(\dfrac{a_1}{1 - \frac{3}{4}} = 20\)
\(\dfrac{a_1}{\frac{1}{4}} = 20\)
\(a_1 = 20 \times \dfrac{1}{4} = \mathbf{5}\) ✅
⭐⭐⭐ Hard
16
Mixed: Arithmetic & Geometric
These are geometric → \((1+3d)^2 = (1+d)(1+7d)\)
\(1+6d+9d^2 = 1+8d+7d^2\)
\(2d^2 - 2d = 0 \Rightarrow d = 1\)
Geometric terms: \(2, 4, 8\) → \(r = \mathbf{2}\) ✅
The 2nd, 4th, and 8th terms of an arithmetic sequence are the 1st, 2nd, and 3rd terms of a geometric sequence.
If the arithmetic sequence has first term \(1\) and common difference \(d\), find the common ratio of the geometric sequence.
If the arithmetic sequence has first term \(1\) and common difference \(d\), find the common ratio of the geometric sequence.
💡 Arithmetic terms: \(a_2=1+d,\ a_4=1+3d,\ a_8=1+7d\)
📝 Explanation
Arithmetic: \(a_2 = 1+d,\ a_4 = 1+3d,\ a_8 = 1+7d\)These are geometric → \((1+3d)^2 = (1+d)(1+7d)\)
\(1+6d+9d^2 = 1+8d+7d^2\)
\(2d^2 - 2d = 0 \Rightarrow d = 1\)
Geometric terms: \(2, 4, 8\) → \(r = \mathbf{2}\) ✅
⭐⭐⭐ Hard
17
Product of Terms
So sequence: \(2, 6, 18, 54, 162\)
Product: \(2 \times 6 \times 18 \times 54 \times 162 = \mathbf{11664}\) ✅
(Or: \(18^5 / \text{...}\) → product \(= 18^5 \div (9 \times 3) = ...\) just multiply directly)
📎 Secret formula: In geo sequence, product of n terms = \((a_1 \cdot a_n)^{n/2}\). Middle term is geometric mean of all!
A geometric sequence has 5 terms. The first term is \(2\) and the fifth term is \(162\).
Find the product of all 5 terms.
Find the product of all 5 terms.
📝 Explanation
Middle (3rd) term = \(\sqrt{a_1 \cdot a_5} = \sqrt{2 \times 162} = \sqrt{324} = 18\)So sequence: \(2, 6, 18, 54, 162\)
Product: \(2 \times 6 \times 18 \times 54 \times 162 = \mathbf{11664}\) ✅
(Or: \(18^5 / \text{...}\) → product \(= 18^5 \div (9 \times 3) = ...\) just multiply directly)
⭐⭐⭐ Hard
18
Exponential Growth — Real World
⚠️ "After 8 hours" means multiplied by r exactly 8 times → \(r^8\), not \(r^7\)!
A population of bacteria doubles every hour. If there are initially \(500\) bacteria,
how many are there after \(8\) hours?
how many are there after \(8\) hours?
💡 \(a_1=500,\ r=2\). After 8 hours = \(a_9\)! (n hours later = term number n+1)
📝 Explanation
After 8 hours: \(500 \times 2^8 = 500 \times 256 = \mathbf{128{,}000}\) ✅⚠️ "After 8 hours" means multiplied by r exactly 8 times → \(r^8\), not \(r^7\)!
⭐⭐⭐ Hard
19
Sum Between Two Terms
\(S_3 = \dfrac{1(1-2^3)}{1-2} = 7\)
Sum = \(S_8 - S_3 = 255 - 7 = \mathbf{248}\) ✅
Check: \(8+16+32+64+128 = 248\) ✔
Find the sum of the geometric series from the 4th term to the 8th term:
\[a_1 = 1,\quad r = 2\] \[\text{Find }\ a_4 + a_5 + a_6 + a_7 + a_8\]
\[a_1 = 1,\quad r = 2\] \[\text{Find }\ a_4 + a_5 + a_6 + a_7 + a_8\]
💡 TRICK: \(S_8 - S_3\) = sum from 4th to 8th
📝 Explanation
\(S_8 = \dfrac{1(1-2^8)}{1-2} = 255\)\(S_3 = \dfrac{1(1-2^3)}{1-2} = 7\)
Sum = \(S_8 - S_3 = 255 - 7 = \mathbf{248}\) ✅
Check: \(8+16+32+64+128 = 248\) ✔
⭐⭐⭐ Hard
20
🎯 Boss Level: Find n for a Given Sum
\(2^n > 1001\)
\(2^9 = 512\) ✗ (not enough)
\(2^{10} = 1024 > 1001\) ✅
So we need \(n = \mathbf{10}\) terms. \(S_{10} = 1023 > 1000\) ✔
How many terms of the geometric series \(1 + 2 + 4 + 8 + \cdots\) must be taken
so that the sum exceeds \(1000\)?
so that the sum exceeds \(1000\)?
💡 Set \(S_n > 1000\), solve: \(2^n - 1 > 1000\) → \(2^n > 1001\)
📝 Explanation
\(S_n = 2^n - 1 > 1000\)\(2^n > 1001\)
\(2^9 = 512\) ✗ (not enough)
\(2^{10} = 1024 > 1001\) ✅
So we need \(n = \mathbf{10}\) terms. \(S_{10} = 1023 > 1000\) ✔