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1.8 Arithmetic & Geometric Sequences, Series & Word Patterns
Ateneo College Entrance Test ACET
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1 Chapter 1: Mathematics
1.1 Real Numbers, Ratios, Rates & Percentages 1.2 Algebraic Expressions, Polynomials & Linear Equations 1.3 Functions, Domain/Range, Graphs & Transformations 1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations 1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area 1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections 1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities 1.8 Arithmetic & Geometric Sequences, Series & Word Patterns 1.9 Permutations, Combinations, Probability & Statistics
2 Chapter 2: English Proficiency
2.1 Subject-Verb Agreement, Verb Tenses & Aspect 2.2 Pronoun-Antecedent Agreement, Case, Misplaced & Dangling Modifiers 2.3 Sentence Structure, Run-Ons, Fragments & Parallelism 2.4 Contextual Vocabulary, Word Roots, Prefixes & Suffixes 2.5 Advanced Synonyms, Antonyms & Nuanced Vocabulary 2.6 Verbal Analogies: Relationship Types & Solving Strategies 2.7 Punctuation, Capitalization, Mechanics & Common Errors 2.8 Diction, Idiomatic Expressions, Prepositions & Frequently Confused Words
3 Chapter 3: Reading Comprehension
3.1 Identifying Main Ideas, Central Themes & Paragraph Summaries 3.2 Locating Explicit Details, Supporting Evidence & Data 3.3 Drawing Logical Inferences & Unstated Conclusions 3.4 Analyzing Author's Purpose, Point of View, Tone & Attitude 3.5 Interpreting Figurative Language, Tone Words & Context Clues 3.6 Text Structure, Rhetorical Devices & Evaluating Arguments
4 Chapter 4: Logical and Abstract Reasoning
4.1 Number & Letter Series Patterns & Formulas 4.2 Abstract Reasoning: Matrix Patterns & Figure Rotations 4.3 Formal Syllogisms, Categorical Logic & Deductive Proofs 4.4 Coding-Decoding, Cipher Patterns & Symbol Operations 4.5 Spatial Visualization, Paper Folding & Venn Diagram Logic
5 Chapter 5: General Knowledge
5.1 Philippine History, Government, Constitution & Civics 5.2 Integrated Science: Biology, Chemistry, Physics & Earth Science 5.3 Philippine & World Geography, Basic Economics & Cultural Facts 5.4 Current Events, World Landmarks, Literature & General Trivia
6 Chapter 6: ACET Essay Writing & Exam Strategies
6.1 ACET Essay: Thesis Development, Structuring & Argumentation 6.2 Time Allocation, Pacing Strategies & Subtest Tactics 6.3 Guessing Strategies (No Penalty), Test Anxiety & Exam Day Readiness
1.8 Arithmetic & Geometric Sequences, Series & Word Patterns
Key Takeaways Explain This Topic AI
Differentiate arithmetic sequences (constant common difference d) from geometric sequences (constant common ratio r).
Apply summation formulas for arithmetic series Sn = n/2 * (a1 + an) and geometric series Sn = a1*(1-r^n)/(1-r).
Evaluate infinite geometric series using S_infinity = a1 / (1-r) when |r| < 1.
Convert repeating decimals to exact rational fractions using infinite series summation rules.
Last updated: July 2026
Arithmetic & Geometric Sequences, Series & Word Patterns
Sequence and series problems on the ACET evaluate term calculation speed, summation formula usage, and pattern recognition.
1. Arithmetic Sequences & Series
An arithmetic sequence has a constant common difference d = a n − a n − 1 d = a_n - a_{n-1} d = a n − a n − 1 . n -th Term Formula: a n = a 1 + ( n − 1 ) d \mathbf{n\text{-th Term Formula:}} \quad a_n = a_1 + (n - 1) d n -th Term Formula: a n = a 1 + ( n − 1 ) d Sum of First n Terms (Series): S n = n 2 ( a 1 + a n ) = n 2 [ 2 a 1 + ( n − 1 ) d ] \textbf{Sum of First } n \textbf{ Terms (Series):} \quad S_n = \frac{n}{2} (a_1 + a_n) = \frac{n}{2} \left[ 2a_1 + (n - 1)d \right] Sum of First n Terms (Series): S n = 2 n ( a 1 + a n ) = 2 n [ 2 a 1 + ( n − 1 ) d ] Arithmetic Mean between a and b : A M = a + b 2 \textbf{Arithmetic Mean between } a \textbf{ and } b \textbf{:} \quad A_{M} = \frac{a + b}{2} Arithmetic Mean between a and b : A M = 2 a + b
ACET Speed Tactic (Sum of Integers): The sum of the first n n n positive integers is S n = n ( n + 1 ) 2 S_n = \frac{n(n + 1)}{2} S n = 2 n ( n + 1 ) .
2. Geometric Sequences & Series
A geometric sequence has a constant common ratio r = a n a n − 1 r = \frac{a_n}{a_{n-1}} r = a n − 1 a n . n -th Term Formula: a n = a 1 ⋅ r n − 1 \mathbf{n\text{-th Term Formula:}} \quad a_n = a_1 \cdot r^{n-1} n -th Term Formula: a n = a 1 ⋅ r n − 1 Sum of First n Terms (Finite Series): S n = a 1 ( 1 − r n ) 1 − r ( r ≠ 1 ) \textbf{Sum of First } n \textbf{ Terms (Finite Series):} \quad S_n = \frac{a_1 (1 - r^n)}{1 - r} \quad (r \neq 1) Sum of First n Terms (Finite Series): S n = 1 − r a 1 ( 1 − r n ) ( r = 1 ) Geometric Mean between a and b : G M = a ⋅ b \textbf{Geometric Mean between } a \textbf{ and } b \textbf{:} \quad G_{M} = \sqrt{a \cdot b} Geometric Mean between a and b : G M = a ⋅ b
3. Infinite Geometric Series & Convergent Applications
An infinite geometric series a 1 + a 1 r + a 1 r 2 + … a_1 + a_1 r + a_1 r^2 + \dots a 1 + a 1 r + a 1 r 2 + … converges to a finite sum if and only if ∣ r ∣ < 1 |r| < 1 ∣ r ∣ < 1 : S ∞ = a 1 1 − r ( ∣ r ∣ < 1 ) S_\infty = \frac{a_1}{1 - r} \quad (|r| < 1) S ∞ = 1 − r a 1 ( ∣ r ∣ < 1 )
Converting Repeating Decimals to Fractions
To express 0. 36 ‾ = 0.363636 … 0.\overline{36} = 0.363636\dots 0. 36 = 0.363636 … as a fraction:
First term a 1 = 0.36 = 36 100 a_1 = 0.36 = \frac{36}{100} a 1 = 0.36 = 100 36 .
Common ratio r = 0.01 = 1 100 r = 0.01 = \frac{1}{100} r = 0.01 = 100 1 . S ∞ = 36 / 100 1 − 1 / 100 = 36 / 100 99 / 100 = 36 99 = 4 11 S_\infty = \frac{36/100}{1 - 1/100} = \frac{36/100}{99/100} = \frac{36}{99} = \frac{4}{11} S ∞ = 1 − 1/100 36/100 = 99/100 36/100 = 99 36 = 11 4
4. Special Sequences & Pattern Recognition
Harmonic Sequence: A sequence whose reciprocals form an arithmetic sequence (e.g., 1 2 , 1 5 , 1 8 , 1 11 \frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11} 2 1 , 5 1 , 8 1 , 11 1 ). Harmonic Mean: H M = 2 a b a + b H_M = \frac{2ab}{a+b} H M = a + b 2 ab .
Fibonacci Sequence: Defined recursively as F 1 = 1 , F 2 = 1 , F n = F n − 1 + F n − 2 F_1 = 1, F_2 = 1, F_n = F_{n-1} + F_{n-2} F 1 = 1 , F 2 = 1 , F n = F n − 1 + F n − 2 . Sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...
Triangular Numbers: T n = n ( n + 1 ) 2 T_n = \frac{n(n+1)}{2} T n = 2 n ( n + 1 ) . Sequence: 1, 3, 6, 10, 15, 21, 28...
- Worked Step-by-Step ACET Exam Problems
- Problem 1: Bouncing Ball Infinite Series Distance
A rubber ball is dropped from a height of 80 meters. Each time it hits the ground, it bounces back up to 3 / 4 3/4 3/4 of its previous height. What is the total vertical distance traveled by the ball before coming to rest?
Step 1: Separate downward and upward distances.
Initial drop distance: 80 m 80\text{ m} 80 m .
Subsequent bounces involve both an upward movement and an equal downward drop.
Step 2: Set up the geometric series for bounces.
First bounce height h 1 = 80 × 3 4 = 60 m h_1 = 80 \times \frac{3}{4} = 60\text{ m} h 1 = 80 × 4 3 = 60 m .
Ball travels 60 m 60\text{ m} 60 m up and 60 m 60\text{ m} 60 m down, total 120 m 120\text{ m} 120 m .
The bounce distance series starts with first term a 1 = 2 × 60 = 120 m a_1 = 2 \times 60 = 120\text{ m} a 1 = 2 × 60 = 120 m and ratio r = 3 4 r = \frac{3}{4} r = 4 3 .
Step 3: Calculate the infinite sum of bounces. S bounces = a 1 1 − r = 120 1 − 3 / 4 = 120 1 / 4 = 480 m S_{\text{bounces}} = \frac{a_1}{1 - r} = \frac{120}{1 - 3/4} = \frac{120}{1/4} = 480\text{ m} S bounces = 1 − r a 1 = 1 − 3/4 120 = 1/4 120 = 480 m
Step 4: Add the initial drop height. Total Distance = Initial Drop + S bounces = 80 + 480 = 560 meters \text{Total Distance} = \text{Initial Drop} + S_{\text{bounces}} = 80 + 480 = 560\text{ meters} Total Distance = Initial Drop + S bounces = 80 + 480 = 560 meters
ACET Shortcut Formula: For drop height H H H and bounce fraction f = a b f = \frac{a}{b} f = b a : Total Distance = H ( 1 + f 1 − f ) = 80 ( 1 + 3 / 4 1 − 3 / 4 ) = 80 ( 7 / 4 1 / 4 ) = 80 × 7 = 560 m \text{Total Distance} = H \left( \frac{1 + f}{1 - f} \right) = 80 \left( \frac{1 + 3/4}{1 - 3/4} \right) = 80 \left( \frac{7/4}{1/4} \right) = 80 \times 7 = 560\text{ m} Total Distance = H ( 1 − f 1 + f ) = 80 ( 1 − 3/4 1 + 3/4 ) = 80 ( 1/4 7/4 ) = 80 × 7 = 560 m
Problem 2: Arithmetic Series Sum of Multiples
Find the sum of all three-digit integers that are exactly divisible by 7.
Step 1: Identify first and last three-digit multiples of 7.
Smallest 3-digit number is 100 100 100 . 100 ÷ 7 = 14.28 ⟹ 15 × 7 = 105 = a 1 100 \div 7 = 14.28 \implies 15 \times 7 = 105 = a_1 100 ÷ 7 = 14.28 ⟹ 15 × 7 = 105 = a 1 .
Largest 3-digit number is 999 999 999 . 999 ÷ 7 = 142.71 ⟹ 142 × 7 = 994 = a n 999 \div 7 = 142.71 \implies 142 \times 7 = 994 = a_n 999 ÷ 7 = 142.71 ⟹ 142 × 7 = 994 = a n .
Step 2: Find total number of terms n n n . a n = a 1 + ( n − 1 ) d ⟹ 994 = 105 + ( n − 1 ) 7 a_n = a_1 + (n - 1)d \implies 994 = 105 + (n - 1)7 a n = a 1 + ( n − 1 ) d ⟹ 994 = 105 + ( n − 1 ) 7 889 = ( n − 1 ) 7 ⟹ n − 1 = 127 ⟹ n = 128 889 = (n - 1)7 \implies n - 1 = 127 \implies n = 128 889 = ( n − 1 ) 7 ⟹ n − 1 = 127 ⟹ n = 128
Step 3: Calculate total sum S 128 S_{128} S 128 . S n = n 2 ( a 1 + a n ) = 128 2 ( 105 + 994 ) = 64 × 1099 = 70 , 336 S_n = \frac{n}{2} (a_1 + a_n) = \frac{128}{2} (105 + 994) = 64 \times 1099 = 70,336 S n = 2 n ( a 1 + a n ) = 2 128 ( 105 + 994 ) = 64 × 1099 = 70 , 336
ACET Speed Tactics & Traps Summary
Problem Type Common Student Trap ACET Speed Tactic
Arithmetic Sum Counting terms incorrectly ( n = last − first n = \text{last} - \text{first} n = last − first ) Use n = a n − a 1 d + 1 n = \frac{a_n - a_1}{d} + 1 n = d a n − a 1 + 1
Infinite Series Applying S ∞ = a 1 1 − r S_\infty = \frac{a_1}{1-r} S ∞ = 1 − r a 1 when ∣ r ∣ ≥ 1 \vert r\vert\ge 1 ∣ r ∣ ≥ 1 Verify ∣ r ∣ < 1 \vert r\vert< 1 ∣ r ∣ < 1 ; if ∣ r ∣ ≥ 1 \vert r\vert\ge 1 ∣ r ∣ ≥ 1 series diverges
Bouncing Ball Forgetting that bounces travel both up and down Use shortcut formula H ⋅ 1 + f 1 − f H \cdot \frac{1+f}{1-f} H ⋅ 1 − f 1 + f
Repeating Decimal Dividing repeating digits by 100 instead of 99 0. a b ‾ = a b 99 0.\overline{ab} = \frac{ab}{99} 0. ab = 99 ab