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4.1 Arithmetic and Geometric Sequences and Series

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Adamson University College Entrance Test AdUCET

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1 Introduction & Exam Overview

1.1 AdUCET Overview, Format, and Admission Process 1.2 Test-Taking and Time-Management Strategies

2 Chapter 2: Mathematics I — Numbers, Algebra, and Functions

2.1 Number Sense, Operations, and Fraction–Decimal–Percent Conversions 2.2 Algebra, Polynomials, and Linear/Quadratic Equations 2.3 Inequalities, Absolute Value, and Systems of Equations 2.4 Radicals, Rational Expressions, and Scientific Notation 2.5 Coordinate Geometry, Lines, and Functions

3 Chapter 3: Mathematics II — Geometry, Measurement, and Trigonometry

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles 3.2 Solid Mensuration, Measurement, and Unit Conversion 3.3 Right-Triangle Trigonometry and Applications

4 Chapter 4: Mathematics III — Sequences, Proportion, and Data

4.1 Arithmetic and Geometric Sequences and Series 4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems 4.3 Descriptive Statistics and Data Interpretation 4.4 Probability: Rules, Compound Events, and Conditional Probability 4.5 Counting Principles: Permutations and Combinations

5 Chapter 5: English I — Reading Comprehension and Vocabulary

5.1 Reading Comprehension: Main Idea, Purpose, and Passage Structure 5.2 Inference, Supporting Detail, and Author's Tone 5.3 Word Roots, Prefixes, and Suffixes 5.4 Vocabulary in Context, Synonyms, Antonyms, and Confusables

6 Chapter 6: English II — Grammar, Usage, and Written Expression

6.1 Subject-Verb Agreement and Verb Tenses 6.2 Pronouns, Modifiers, and Parallel Structure 6.3 Sentence Boundaries, Clauses, and Punctuation 6.4 Sentence Correction, Redundancy, and Idiomatic Usage 6.5 Verbal Analogies and Relationship Bridges 6.6 Paragraph Organization, Coherence, and Transitions

7 Chapter 7: General Science I — Biology and Chemistry

7.1 Cell Biology, Transport, and Cellular Energetics 7.2 Genetics, Heredity, and Molecular Biology 7.3 Human Body Systems, Ecology, and Evolution 7.4 Matter, Phase Changes, and Atomic Structure 7.5 Periodic Trends, Chemical Bonding, Reactions, Acids and Bases

8 Chapter 8: General Science II — Physics and Earth Science

8.1 Mechanics: Motion, Newton's Laws, Work, and Energy 8.2 Waves, Sound, Light, Heat, Electricity, and Magnetism 8.3 Earth's Structure, Plate Tectonics, and the Rock Cycle 8.4 The Atmosphere, Weather, the Water Cycle, and Astronomy

9 Chapter 9: Aptitude and Abstract Reasoning

9.1 Figure Series and Pattern Progression 9.2 Matrix Reasoning and Figural Analogies 9.3 Spatial Reasoning: Rotation, Reflection, and Chirality 9.4 Paper Folding, Cube Nets, and Orthographic Views 9.5 Deductive Reasoning, Syllogisms, and Venn Diagrams 9.6 Number, Letter, and Alphanumeric Series 9.7 Odd-One-Out, Classification, and Logic Puzzles

4.1 Arithmetic and Geometric Sequences and Series

Key Takeaways Explain This Topic AI

The nth term of an arithmetic sequence is a_n = a_1 + (n-1)d, and the sum of the first n terms is S_n = (n/2)(a_1 + a_n).

A geometric sequence uses a_n = a_1 r^(n-1), and its first n terms total S_n = a_1(1 - r^n)/(1 - r) whenever r is not 1.

An infinite geometric series converges to a_1/(1 - r) only when |r| < 1; if |r| is 1 or more the series diverges and has no finite sum.

A fixed amount added each period is arithmetic; a fixed percentage of the current value is geometric, with r = 1 + rate for growth and r = 1 - rate for depreciation.

Inserting k arithmetic means gives d = (last - first)/(k + 1), while inserting k geometric means gives r^(k+1) = last divided by first.

Last updated: August 2026

4.1 Arithmetic and Geometric Sequences and Series

A sequence is an ordered list of numbers called terms , written a 1 , a 2 , a 3 , … , a n a_1, a_2, a_3, \dots, a_n a 1 ​ , a 2 ​ , a 3 ​ , … , a n ​ , where the subscript records the term's position. A series is what you get when you add those terms together. Entrance-test items on this topic almost never ask you to prove anything; they hand you a realistic pattern — bleacher rows, a salary ladder, a depreciating jeepney, splitting bacteria — and ask for one particular term, for a running total, or for the number of terms. Every one of those items collapses to a single decision: is the pattern built by adding the same amount each step, or by multiplying by the same factor each step?

1. Arithmetic sequences

An arithmetic sequence adds a fixed number, the common difference d d d , to move from any term to the next: d = a n − a n − 1 d = a_n - a_{n-1} d = a n ​ − a n − 1 ​

In 7 , 11 , 15 , 19 , … 7, 11, 15, 19, \dots 7 , 11 , 15 , 19 , … every gap is 4 4 4 , so d = 4 d = 4 d = 4 . A negative d d d makes the list fall: 50 , 44 , 38 , … 50, 44, 38, \dots 50 , 44 , 38 , … has d = − 6 d = -6 d = − 6 .

The n n n -th term. a n = a 1 + ( n − 1 ) d a_n = a_1 + (n-1)d a n ​ = a 1 ​ + ( n − 1 ) d

The multiplier is ( n − 1 ) (n-1) ( n − 1 ) , not n n n , because the first term needs no step at all. For 7 , 11 , 15 , … 7, 11, 15, \dots 7 , 11 , 15 , … the twentieth term is a 20 = 7 + 19 ( 4 ) = 83 a_{20} = 7 + 19(4) = 83 a 20 ​ = 7 + 19 ( 4 ) = 83 .

Locating a known value. Set a n a_n a n ​ equal to the value and solve for n n n . Is 143 143 143 in the list above? 143 = 7 + ( n − 1 ) 4 ⇒ 136 = 4 ( n − 1 ) ⇒ n = 35 143 = 7 + (n-1)4 \Rightarrow 136 = 4(n-1) \Rightarrow n = 35 143 = 7 + ( n − 1 ) 4 ⇒ 136 = 4 ( n − 1 ) ⇒ n = 35 , a positive whole number, so yes — it is the 35th term. A fractional n n n would mean the value is not a term at all.

Sum of the first n n n terms. S n = n 2 ( a 1 + a n ) = n 2 [ 2 a 1 + ( n − 1 ) d ] S_n = \frac{n}{2}\left(a_1 + a_n\right) = \frac{n}{2}\left[2a_1 + (n-1)d\right] S n ​ = 2 n ​ ( a 1 ​ + a n ​ ) = 2 n ​ [ 2 a 1 ​ + ( n − 1 ) d ]

Use the first form when the last term is already known, the second when you know only a 1 a_1 a 1 ​ , d d d and n n n .

Arithmetic means are the terms inserted between two given numbers so that the whole list stays arithmetic. Inserting k k k means produces k + 2 k+2 k + 2 terms and therefore k + 1 k+1 k + 1 gaps, so d = last − first k + 1 d = \dfrac{\text{last} - \text{first}}{k+1} d = k + 1 last − first ​ . Insert three arithmetic means between 7 7 7 and 27 27 27 : d = 27 − 7 4 = 5 d = \frac{27-7}{4} = 5 d = 4 27 − 7 ​ = 5 , giving 7 , 12 , 17 , 22 , 27 7, 12, 17, 22, 27 7 , 12 , 17 , 22 , 27 . A single arithmetic mean is just the ordinary average a + b 2 \frac{a+b}{2} 2 a + b ​ .

2. Geometric sequences

A geometric sequence multiplies by a fixed non-zero number, the common ratio r r r : r = a n a n − 1 r = \frac{a_n}{a_{n-1}} r = a n − 1 ​ a n ​ ​

In 6 , 18 , 54 , 162 , … 6, 18, 54, 162, \dots 6 , 18 , 54 , 162 , … each term is triple the one before it, so r = 3 r = 3 r = 3 and a 7 = 6 ⋅ 3 6 = 6 ( 729 ) = 4,374 a_7 = 6 \cdot 3^{6} = 6(729) = 4{,}374 a 7 ​ = 6 ⋅ 3 6 = 6 ( 729 ) = 4 , 374 .

Sum of the first n n n terms. S n = a 1 ( 1 − r n ) 1 − r , r ≠ 1 S_n = \frac{a_1\left(1 - r^{n}\right)}{1 - r}, \qquad r \neq 1 S n ​ = 1 − r a 1 ​ ( 1 − r n ) ​ , r  = 1

For 6 , 18 , 54 , … 6, 18, 54, \dots 6 , 18 , 54 , … the first six terms total S 6 = 6 ( 1 − 729 ) 1 − 3 = 6 ( − 728 ) − 2 = 2,184 S_6 = \frac{6(1 - 729)}{1 - 3} = \frac{6(-728)}{-2} = 2{,}184 S 6 ​ = 1 − 3 6 ( 1 − 729 ) ​ = − 2 6 ( − 728 ) ​ = 2 , 184 , which you can check by hand: 6 + 18 + 54 + 162 + 486 + 1458 = 2,184 6+18+54+162+486+1458 = 2{,}184 6 + 18 + 54 + 162 + 486 + 1458 = 2 , 184 .

Geometric means follow the same logic as arithmetic means but with multiplication: inserting k k k geometric means creates k + 1 k+1 k + 1 jumps, so r k + 1 = last first r^{\,k+1} = \dfrac{\text{last}}{\text{first}} r k + 1 = first last ​ . Insert two geometric means between 5 5 5 and 135 135 135 : r 3 = 27 r^3 = 27 r 3 = 27 , so r = 3 r = 3 r = 3 and the list is 5 , 15 , 45 , 135 5, 15, 45, 135 5 , 15 , 45 , 135 . The single geometric mean of a a a and b b b is ± a b \pm\sqrt{ab} ± ab ​ ; for 4 4 4 and 25 25 25 it is ± 10 \pm 10 ± 10 .

3. Infinite geometric series

Add a geometric sequence forever and exactly one of two things happens. If ∣ r ∣ ≥ 1 \lvert r \rvert \ge 1 ∣ r ∣ ≥ 1 the terms never shrink and the total runs away — the series diverges and has no finite sum. If ∣ r ∣ < 1 \lvert r \rvert < 1 ∣ r ∣ < 1 each term is a fraction of the one before, r n r^n r n collapses toward zero, and the running total settles on a ceiling: S ∞ = a 1 1 − r , ∣ r ∣ < 1 S_\infty = \frac{a_1}{1 - r}, \qquad \lvert r \rvert < 1 S ∞ ​ = 1 − r a 1 ​ ​ , ∣ r ∣ < 1

So 24 + 8 + 8 3 + … 24 + 8 + \frac{8}{3} + \dots 24 + 8 + 3 8 ​ + … has r = 1 3 r = \frac13 r = 3 1 ​ and S ∞ = 24 1 − 1 / 3 = 24 2 / 3 = 36 S_\infty = \frac{24}{1 - 1/3} = \frac{24}{2/3} = 36 S ∞ ​ = 1 − 1/3 24 ​ = 2/3 24 ​ = 36 , while 2 + 6 + 18 + … 2 + 6 + 18 + \dots 2 + 6 + 18 + … has r = 3 r = 3 r = 3 and simply has no sum. Examiners like to offer a tidy number for a divergent series, so test ∣ r ∣ \lvert r \rvert ∣ r ∣ before computing anything.

Repeating decimals as fractions

A repeating decimal is an infinite geometric series. 0. 36 ‾ = 0.36 + 0.0036 + … 0.\overline{36} = 0.36 + 0.0036 + \dots 0. 36 = 0.36 + 0.0036 + … has a 1 = 0.36 a_1 = 0.36 a 1 ​ = 0.36 and r = 0.01 r = 0.01 r = 0.01 , so S ∞ = 0.36 0.99 = 36 99 = 4 11 S_\infty = \frac{0.36}{0.99} = \frac{36}{99} = \frac{4}{11} S ∞ ​ = 0.99 0.36 ​ = 99 36 ​ = 11 4 ​ . That generalises into a shortcut: a block of k k k repeating digits sitting immediately after the decimal point equals that block over k k k nines — 0. 5 ‾ = 5 9 0.\overline{5} = \frac59 0. 5 = 9 5 ​ and 0. 123 ‾ = 123 999 = 41 333 0.\overline{123} = \frac{123}{999} = \frac{41}{333} 0. 123 = 999 123 ​ = 333 41 ​ . When a non-repeating digit comes first, peel it off and treat the rest as a series: 0.1 6 ‾ = 0.1 + ( 0.06 + 0.006 + … ) = 1 10 + 0.06 0.9 = 1 10 + 1 15 = 1 6 0.1\overline{6} = 0.1 + \left(0.06 + 0.006 + \dots\right) = \frac{1}{10} + \frac{0.06}{0.9} = \frac{1}{10} + \frac{1}{15} = \frac{1}{6} 0.1 6 = 0.1 + ( 0.06 + 0.006 + … ) = 10 1 ​ + 0.9 0.06 ​ = 10 1 ​ + 15 1 ​ = 6 1 ​ .

4. Harmonic and recursive sequences

A harmonic sequence is one whose reciprocals form an arithmetic sequence. In 1 2 , 1 5 , 1 8 , 1 11 , … \frac12, \frac15, \frac18, \frac1{11}, \dots 2 1 ​ , 5 1 ​ , 8 1 ​ , 11 1 ​ , … the denominators 2 , 5 , 8 , 11 2, 5, 8, 11 2 , 5 , 8 , 11 climb by 3 3 3 , so the next reciprocal is 14 14 14 and the next term is 1 14 \frac1{14} 14 1 ​ . Never hunt for a common difference among the harmonic terms themselves — flip every term first. The harmonic mean of a a a and b b b is 2 a b a + b \frac{2ab}{a+b} a + b 2 ab ​ ; for 4 4 4 and 12 12 12 that is 96 16 = 6 \frac{96}{16} = 6 16 96 ​ = 6 .

A recursive (Fibonacci-style) sequence defines each term from the ones before it, most often a n = a n − 1 + a n − 2 a_n = a_{n-1} + a_{n-2} a n ​ = a n − 1 ​ + a n − 2 ​ . Seeded with 3 3 3 and 4 4 4 it runs 3 , 4 , 7 , 11 , 18 , 29 , 47 3, 4, 7, 11, 18, 29, 47 3 , 4 , 7 , 11 , 18 , 29 , 47 . There is no common difference and no common ratio, so whenever both tests fail, check whether each term is the sum (or difference, or product) of its predecessors and simply extend the list.

5. Choosing the right model

  • Feature Arithmetic Geometric
  • Built by adding a constant d d d multiplying by a constant r r r
  • Test for it consecutive differences are equal consecutive ratios are equal
  • n n n -th term a 1 + ( n − 1 ) d a_1 + (n-1)d a 1 ​ + ( n − 1 ) d a 1 r n − 1 a_1 r^{\,n-1} a 1 ​ r n − 1

Sum of n n n terms n 2 ( a 1 + a n ) \frac{n}{2}(a_1 + a_n) 2 n ​ ( a 1 ​ + a n ​ ) a 1 ( 1 − r n ) 1 − r \frac{a_1(1 - r^n)}{1 - r} 1 − r a 1 ​ ( 1 − r n ) ​

  • Infinite sum none a 1 1 − r \frac{a_1}{1-r} 1 − r a 1 ​ ​ , only when ∣ r ∣ < 1 \lvert r \rvert < 1 ∣ r ∣ < 1
  • Plotted points fall on a straight line fall on an exponential curve
  • Typical wording "adds ₱500 every year" "grows 5% every year"

Word problems announce their model in one phrase:

  • Scenario Signal in the wording Model Parameter
  • Covered-court bleachers, each row 3 seats more than the last fixed extra amount arithmetic d = 3 d = 3 d = 3
  • Salary rises by a flat ₱1,200 a year fixed peso raise arithmetic d = 1,200 d = 1{,}200 d = 1 , 200
  • Salary rises by 8% a year percent of the current value geometric r = 1.08 r = 1.08 r = 1.08

Ball rebounds to 3 5 \frac35 5 3 ​ of its previous height fixed fraction of the previous geometric, infinite r = 0.6 r = 0.6 r = 0.6

  • Tricycle loses 15% of its value yearly percent of value lost geometric r = 0.85 r = 0.85 r = 0.85
  • Bacteria double every 20 minutes doubling geometric r = 2 r = 2 r = 2
  • Each term is the sum of the two before it the rule refers to itself recursive two seed terms

Terms are 1 3 , 1 7 , 1 11 \frac13, \frac17, \frac1{11} 3 1 ​ , 7 1 ​ , 11 1 ​ reciprocals look regular harmonic flip first

6. Worked example 1 — tiered seating

Problem. The bleachers of a barangay covered court in Sampaloc have 18 rows. The front row seats 14 people, and each row behind it seats 3 more than the row in front. (a) How many seats are in the last row? (b) How many seats are there altogether? (c) Which is the first row that seats more than 50 people?

Each row adds a fixed 3 seats, so the rows are arithmetic with a 1 = 14 a_1 = 14 a 1 ​ = 14 , d = 3 d = 3 d = 3 , n = 18 n = 18 n = 18 .

(a) a 18 = 14 + ( 18 − 1 ) ( 3 ) = 14 + 51 = 65 a_{18} = 14 + (18-1)(3) = 14 + 51 = 65 a 18 ​ = 14 + ( 18 − 1 ) ( 3 ) = 14 + 51 = 65 seats in the last row.

(b) S 18 = 18 2 ( 14 + 65 ) = 9 ( 79 ) = 711 S_{18} = \frac{18}{2}(14 + 65) = 9(79) = 711 S 18 ​ = 2 18 ​ ( 14 + 65 ) = 9 ( 79 ) = 711 seats in total.

(c) Solve 14 + 3 ( n − 1 ) > 50 ⇒ 3 ( n − 1 ) > 36 ⇒ n > 13 14 + 3(n-1) > 50 \Rightarrow 3(n-1) > 36 \Rightarrow n > 13 14 + 3 ( n − 1 ) > 50 ⇒ 3 ( n − 1 ) > 36 ⇒ n > 13 . The first whole-number row is n = 14 n = 14 n = 14 , seating 14 + 39 = 53 14 + 39 = 53 14 + 39 = 53 . Row 13 seats exactly 50, which is not "more than 50" — that strict-inequality boundary is the trap.

7. Worked example 2 — depreciation and a pledge ceiling

Problem. (a) A second-hand jeepney bought for ₱480,000 loses 25% of its current value every year. What is it worth at the start of year 5, and how much value has it lost in total? (b) A parish fundraising drive collects ₱64,000 on its first day, and on each following day collects three-quarters of the previous day's amount. How much comes in over the first three days, and what total can the drive never exceed?

(a) Losing 25% means keeping 75%, so r = 0.75 r = 0.75 r = 0.75 — not 0.25 0.25 0.25 .

The start of year 5 is four multiplications after purchase: 480,000 → 360,000 → 270,000 → 202,500 → 151,875. The jeepney is worth ₱151,875.

Total value lost = 480,000 − 151,875 = 328,125 = 480{,}000 - 151{,}875 = 328{,}125 = 480 , 000 − 151 , 875 = 328 , 125 , so ₱328,125. Cross-check with the yearly losses: 120,000 + 90,000 + 67,500 + 50,625 = 328,125 120{,}000 + 90{,}000 + 67{,}500 + 50{,}625 = 328{,}125 120 , 000 + 90 , 000 + 67 , 500 + 50 , 625 = 328 , 125 . ✓

(b) Daily collections are geometric with a 1 = 64,000 a_1 = 64{,}000 a 1 ​ = 64 , 000 and r = 3 4 r = \frac34 r = 4 3 ​ .

First three days: 64,000 + 48,000 + 36,000 = 148,000 64{,}000 + 48{,}000 + 36{,}000 = 148{,}000 64 , 000 + 48 , 000 + 36 , 000 = 148 , 000 , i.e. ₱148,000. The formula agrees: S 3 = 64,000 ( 1 − 27 64 ) 1 − 3 4 = 64,000 ⋅ 37 64 ⋅ 4 = 148,000 S_3 = \frac{64{,}000\left(1 - \frac{27}{64}\right)}{1 - \frac34} = 64{,}000 \cdot \frac{37}{64} \cdot 4 = 148{,}000 S 3 ​ = 1 − 4 3 ​ 64 , 000 ( 1 − 64 27 ​ ) ​ = 64 , 000 ⋅ 64 37 ​ ⋅ 4 = 148 , 000 .

Since ∣ r ∣ = 0.75 < 1 \lvert r \rvert = 0.75 < 1 ∣ r ∣ = 0.75 < 1 , the infinite total converges: S ∞ = 64,000 1 − 0.75 = 64,000 0.25 = 256,000 S_\infty = \frac{64{,}000}{1 - 0.75} = \frac{64{,}000}{0.25} = 256{,}000 S ∞ ​ = 1 − 0.75 64 , 000 ​ = 0.25 64 , 000 ​ = 256 , 000 . The drive can creep toward ₱256,000 but can never reach it.

8. Common traps

Off-by-one on n − 1 n-1 n − 1 . The 10th term takes nine steps, not ten. Writing a 1 + n d a_1 + nd a 1 ​ + n d is the single most frequent arithmetic-sequence error.

"In the 10th year" versus "over 10 years." The first asks for a 10 a_{10} a 10 ​ , the second for S 10 S_{10} S 10 ​ . Read the preposition before you pick a formula.

Percent language hides a ratio. "Increases by 6%" means r = 1.06 r = 1.06 r = 1.06 ; "falls by 6%" means r = 0.94 r = 0.94 r = 0.94 . Using r = 0.06 r = 0.06 r = 0.06 turns steady growth into instant collapse.

Divergent series. When ∣ r ∣ ≥ 1 \lvert r \rvert \ge 1 ∣ r ∣ ≥ 1 there is no infinite sum, however inviting the answer choices look.

Bouncing objects. The first drop is travelled once; every rebound height is travelled twice — up and then down again.

Year 0 versus year 1. Decide whether the purchase price is a 1 a_1 a 1 ​ or the value after one year, then keep that choice for the whole item.