How to Solve Operations in a Sequence?

Solving operations in a sequence can seem daunting at first, but with a clear understanding of the rules and a bit of practice, it becomes much easier. Let’s break it down step by step.

Order of Operations

The most important rule to remember when solving operations in a sequence is the order of operations. This is often remembered by the acronym PEMDAS:

  • Parentheses
  • Exponents (or indices)
  • Multiplication and Division (from left to right)
  • Addition and Subtraction (from left to right)

Example

Consider the expression: $8 + 2 times (3^2 – 1)$. Let’s solve it step by step following PEMDAS.

  1. Parentheses: Solve the expression inside the parentheses first.

    $3^2 – 1 = 9 – 1 = 8$

    So, the expression now is: $8 + 2 times 8$

  2. Exponents: There are no exponents left to solve.

  3. Multiplication and Division: Next, perform the multiplication.

    $2 times 8 = 16$

    So, the expression now is: $8 + 16$

  4. Addition and Subtraction: Finally, perform the addition.

    $8 + 16 = 24$

So, the value of the expression $8 + 2 times (3^2 – 1)$ is 24.

Detailed Steps

  1. Parentheses

    Parentheses are used to group parts of an expression that should be evaluated first. Always start with the innermost parentheses and work your way outward.

Example

For the expression $5 times (2 + (3 – 1))$, solve the innermost parentheses first:

$3 – 1 = 2$

So, the expression becomes $5 times (2 + 2)$. Now solve the remaining parentheses:

$2 + 2 = 4$

So, the expression becomes $5 times 4 = 20$

  1. Exponents

    After solving any parentheses, evaluate any exponents next.

Example

For the expression $2^3 + 4$, solve the exponent first:

$2^3 = 8$

So, the expression becomes $8 + 4 = 12$

  1. Multiplication and Division

    Multiplication and division are performed next, from left to right. It’s important to note that these operations are of equal priority, so you should solve them as they appear from left to right.

Example

For the expression $6 times 3 times 2 / 4$, solve from left to right:

$6 times 3 = 18$

$18 times 2 = 36$

$36 / 4 = 9$

  1. Addition and Subtraction

    Finally, perform any addition and subtraction, from left to right. These operations are also of equal priority.

Example

For the expression $10 – 3 + 2$, solve from left to right:

$10 – 3 = 7$

$7 + 2 = 9$

Common Mistakes and How to Avoid Them

Ignoring the Order of Operations

One common mistake is to ignore the order of operations and solve the expression from left to right without following PEMDAS. Always remember to follow the order of operations.

Forgetting to Solve Inside Parentheses First

Another common mistake is to forget to solve the expressions inside parentheses first. Always start with the innermost parentheses and work your way outward.

Incorrectly Solving Multiplication and Division

Multiplication and division should be solved from left to right, not just multiplication first and then division. The same goes for addition and subtraction.

Practice Problems

Here are some practice problems to help you get more comfortable with solving operations in a sequence:

  1. $7 + 3 times (10 – 4)$
  2. $(8 + 2) times (3^2 – 1)$
  3. $5 + 6 times 2^2 – 3$
  4. $4 times (6 + 2) / 2$
  5. $10 / (2 + 3) times 4$

Solutions

  1. $7 + 3 times (10 – 4) = 7 + 3 times 6 = 7 + 18 = 25$
  2. $(8 + 2) times (3^2 – 1) = 10 times 8 = 80$
  3. $5 + 6 times 2^2 – 3 = 5 + 6 times 4 – 3 = 5 + 24 – 3 = 26$
  4. $4 times (6 + 2) / 2 = 4 times 8 / 2 = 32 / 2 = 16$
  5. $10 / (2 + 3) times 4 = 10 / 5 times 4 = 2 times 4 = 8$

Conclusion

Mastering the order of operations is crucial for solving mathematical expressions accurately. By following the PEMDAS rule, you can systematically approach and solve any expression. Practice regularly, and soon it will become second nature.

Citations

  1. 1. Khan Academy – Order of Operations
  2. 2. Math is Fun – Order of Operations
  3. 3. Purplemath – Order of Operations
  4. 4. Mathway – Order of Operations

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ