1.8 Arithmetic & Geometric Sequences, Series & Word Patterns

Key Takeaways

  • 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 $d = a_n - a_{n-1}$.

n-th Term Formula:an=a1+(n1)d\mathbf{n\text{-th Term Formula:}} \quad a_n = a_1 + (n - 1) d SumofFirstnTerms(Series):Sn=n2(a1+an)=n2[2a1+(n1)d]\mathbf{Sum of First } n \mathbf{ Terms (Series):} \quad S_n = \frac{n}{2} (a_1 + a_n) = \frac{n}{2} \left[ 2a_1 + (n - 1)d \right] ArithmeticMeanbetweenaandb:AM=a+b2\mathbf{Arithmetic Mean between } a \mathbf{ and } b \mathbf{:} \quad A_{M} = \frac{a + b}{2}

ACET Speed Tactic (Sum of Integers): The sum of the first $n$ positive integers is $S_n = \frac{n(n + 1)}{2}$.


2. Geometric Sequences & Series

A geometric sequence has a constant common ratio $r = \frac{a_n}{a_{n-1}}$.

n-th Term Formula:an=a1rn1\mathbf{n\text{-th Term Formula:}} \quad a_n = a_1 \cdot r^{n-1} SumofFirstnTerms(FiniteSeries):Sn=a1(1rn)1r(r1)\mathbf{Sum of First } n \mathbf{ Terms (Finite Series):} \quad S_n = \frac{a_1 (1 - r^n)}{1 - r} \quad (r \neq 1) GeometricMeanbetweenaandb:GM=ab\mathbf{Geometric Mean between } a \mathbf{ and } b \mathbf{:} \quad G_{M} = \sqrt{a \cdot b}


3. Infinite Geometric Series & Convergent Applications

An infinite geometric series $a_1 + a_1 r + a_1 r^2 + \dots$ converges to a finite sum if and only if $|r| < 1$:

S=a11r(r<1)S_\infty = \frac{a_1}{1 - r} \quad (|r| < 1)

Converting Repeating Decimals to Fractions

To express $0.\overline{36} = 0.363636\dots$ as a fraction:

  • First term $a_1 = 0.36 = \frac{36}{100}$.
  • Common ratio $r = 0.01 = \frac{1}{100}$. S=36/10011/100=36/10099/100=3699=411S_\infty = \frac{36/100}{1 - 1/100} = \frac{36/100}{99/100} = \frac{36}{99} = \frac{4}{11}

4. Special Sequences & Pattern Recognition

  1. Harmonic Sequence: A sequence whose reciprocals form an arithmetic sequence (e.g., $\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}$). Harmonic Mean: $H_M = \frac{2ab}{a+b}$.
  2. Fibonacci Sequence: Defined recursively as $F_1 = 1, F_2 = 1, F_n = F_{n-1} + F_{n-2}$. Sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...
  3. Triangular Numbers: $T_n = \frac{n(n+1)}{2}$. 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 $3/4$ 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: $80\text{ m}$.
  • Subsequent bounces involve both an upward movement and an equal downward drop.

Step 2: Set up the geometric series for bounces.

  • First bounce height $h_1 = 80 \times \frac{3}{4} = 60\text{ m}$.
  • Ball travels $60\text{ m}$ up and $60\text{ m}$ down, total $120\text{ m}$.
  • The bounce distance series starts with first term $a_1 = 2 \times 60 = 120\text{ m}$ and ratio $r = \frac{3}{4}$.

Step 3: Calculate the infinite sum of bounces. Sbounces=a11r=12013/4=1201/4=480 mS_{\text{bounces}} = \frac{a_1}{1 - r} = \frac{120}{1 - 3/4} = \frac{120}{1/4} = 480\text{ m}

Step 4: Add the initial drop height. Total Distance=Initial Drop+Sbounces=80+480=560 meters\text{Total Distance} = \text{Initial Drop} + S_{\text{bounces}} = 80 + 480 = 560\text{ meters}

ACET Shortcut Formula: For drop height $H$ and bounce fraction $f = \frac{a}{b}$: Total Distance=H(1+f1f)=80(1+3/413/4)=80(7/41/4)=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}


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 $100$. $100 \div 7 = 14.28 \implies 15 \times 7 = 105 = a_1$.
  • Largest 3-digit number is $999$. $999 \div 7 = 142.71 \implies 142 \times 7 = 994 = a_n$.

Step 2: Find total number of terms $n$. an=a1+(n1)d    994=105+(n1)7a_n = a_1 + (n - 1)d \implies 994 = 105 + (n - 1)7 889=(n1)7    n1=127    n=128889 = (n - 1)7 \implies n - 1 = 127 \implies n = 128

Step 3: Calculate total sum $S_{128}$. Sn=n2(a1+an)=1282(105+994)=64×1099=70,336S_n = \frac{n}{2} (a_1 + a_n) = \frac{128}{2} (105 + 994) = 64 \times 1099 = 70,336


ACET Speed Tactics & Traps Summary

Problem TypeCommon Student TrapACET Speed Tactic
Arithmetic SumCounting terms incorrectly ($n = \text{last} - \text{first}$)Use $n = \frac{a_n - a_1}{d} + 1$
Infinite SeriesApplying $S_\infty = \frac{a_1}{1-r}$ when $r
Bouncing BallForgetting that bounces travel both up and downUse shortcut formula $H \cdot \frac{1+f}{1-f}$
Repeating DecimalDividing repeating digits by 100 instead of 99$0.\overline{ab} = \frac{ab}{99}$
Test Your Knowledge

What is the sum of the first 20 terms of an arithmetic sequence whose first term is 5 and common difference is 4?

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Test Your Knowledge

Find the sum of the infinite geometric series: 12 + 4 + 4/3 + 4/9 + ...

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Test Your Knowledge

The 3rd term of a geometric sequence is 18 and the 6th term is 486. What is the first term of the sequence?

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