Consecutive Numbers
In an ordered sequence, any element that directly succeeds the previous one according to a specific rule is called a consecutive element. For instance, integers arranged in succession based on a specific rule are known as consecutive numbers.
- The consecutive term following $n$ is $n + 1$.
- In the context of odd numbers, the consecutive term following $7$ is $9$.
- In the algebraic expression $1+ x + x^2+ x^3$, the consecutive term following $x^2$ is $x^3$.
Numbers that follow one another according to a steady, predefined pattern are called consecutive numbers.
Examples:
Given $n \in \mathbb{Z}$:
- Consecutive integers: $\dots, -3, -2, -1, 0, 1, 2, 3, \dots, n, n+1, \dots$
- Consecutive even numbers: $\dots, -6, -4, -2, 0, 2, 4, 6, \dots, 2n, 2n+2, \dots$
- Consecutive odd numbers: $\dots, -5, -3, -1, 1, 3, 5, \dots, 2n-1, 2n+1, \dots$
- Consecutive integers that are multiples of 3: $\dots, -6, -3, 0, 3, 6, 9, \dots, 3n, 3n+3, \dots$
- Consecutive numbers with the general term $n^3$: $\dots, -27, -8, -1, 0, 1, 8, \dots, n^3, (n+1)^3, \dots$
- Consecutive integers with a constant difference of 5 between terms:
$$\dots, -12, -7, -2, 3, 8, 13, \dots, 5n+3, 5n+8, \dots$$
$$\dots, -14, -9, -4, 1, 6, 11, \dots, 5n+1, 5n+6, \dots$$
Example:
Given that $a < b < c$ are consecutive odd numbers, let us find the value of the following expression: $$(c-a) \cdot (b-c) + (a-c) \cdot (a-b)^3 \cdot (b-c)$$
Since $a$, $b$, and $c$ are consecutive odd numbers, they increase by a constant step of 2. Therefore, if we define the terms as $n$, $n+2$, and $n+4$ respectively:
$$(c-a) \cdot (b-c) = (n+4-n) \cdot [n+2-(n+4)] = 4 \cdot (-2) = -8$$
$$(a-c) \cdot (a-b)^3 \cdot (b-c) = [n-(n+4)] \cdot [n-(n+2)]^3 \cdot [n+2-(n+4)] = (-4) \cdot (-2)^3 \cdot (-2) = (-4) \cdot (-8) \cdot (-2) = -64$$
Combining these two evaluations together:
$$(-8) + (-64) = -72$$
Since the variable $n$ cancels out and does not impact the final result, we can alternative solve this problem by assigning simple numerical values that satisfy the condition, such as:
Choosing $a=1, b=3, c=5$:
$$= (5 – 1) \cdot (3 – 5) + (1 – 5) \cdot (1 – 3)^3 \cdot (3 – 5)$$
$$= 4 \cdot (-2) + (-4) \cdot (-2)^3 \cdot (-2)$$
$$= -8 + (-64) = -72$$
Example:
If the sum of five consecutive odd numbers is equal to $k$, let us express the largest of these numbers in terms of $k$.
Since the difference between any two consecutive odd numbers is 2, if we denote the smallest number as $x$, our five numbers are:
$$x, \quad x+2, \quad x+4, \quad x+6, \quad x+8$$ Since their sum equals $k$:
$$5x + 20 = k \implies x = \frac{k-20}{5} = \frac{k}{5} – 4$$ Therefore, the largest odd number is given by:
$$x+8 = \frac{k}{5} – 4 + 8 \implies x+8 = \frac{k+20}{5}$$
Let’s look at an alternative solution using a second method. In a finite arithmetic sequence with an odd number of terms, the middle term is always exactly equal to the arithmetic mean of the terms. Hence, the middle term of these five numbers, which is $x + 4$, equals $\frac{k}{5}$. Since the largest term is exactly 4 units greater than the middle term:
$$\text{Largest Term} = \frac{k}{5} + 4 \implies \text{Largest Term} = \frac{k+20}{5}$$
QUESTION 7
The product of three consecutive integers is 120. Based on this, what is the sum of these three numbers?
\[ \text{A) } 9 \quad \text{B) } 12 \quad \text{C) } 15 \quad \text{D) } 18 \quad \text{E) } 21 \]
Solution:
Let us decompose the number 120 into groups of three integer factors to check which combination consists of consecutive terms:
$$120 = 1 \cdot 2 \cdot 60 \quad \quad 120 = 2 \cdot 3 \cdot 20 \quad \quad 120 = 3 \cdot 4 \cdot 10 \quad \quad 120 = 4 \cdot 5 \cdot 6$$
The factors that perfectly satisfy the conditions given in the question are $(4, 5, 6)$. Therefore, the sum of these numbers is:
$$4 + 5 + 6 = 15$$
$\textbf{Correct Answer: C}$
Term: In an algebraic expression, each individual quantity separated from others by a plus $(+)$ or minus $(-)$ sign is called a term. A term can stand alone, exist as a product, be enclosed within parentheses, or be structured under a fraction or a root. For example, in the expression:
$$3x^3 + 2x^2y – (2x+y) + \frac{x^2-1}{y+1} – \sqrt[3]{1-x^2}$$ the component quantities
$$3x^3, \quad 2x^2y, \quad -(2x+y), \quad \frac{x^2-1}{y+1}, \quad \text{and} \quad -\sqrt[3]{1-x^2}$$ are all individual terms.
Finite Sums of Consecutive Numbers:
For any arithmetic sequence where the difference between any two consecutive terms is constant, the sum of its terms can be calculated using the following general formula:
$$\text{Sum} = \frac{\textbf{(First Term + Last Term)}}{2} \cdot \textbf{Number of Terms}$$ The total number of terms is found using the formula:
$$\text{Number of Terms} = \frac{\textbf{(Last Term – First Term)}}{\textbf{Common Difference}} + 1$$
To express these equations systematically, let us define: first term = $m$, last term = $n$, common difference = $k$, number of terms = $t$, and total sum = $T$. The formulas rearrange as follows:
$$ \textbf{t} = \frac{\textbf{n – m}}{\textbf{k}} + 1 $$
$$ \textbf{T} = \frac{\textbf{(n + m)}}{\textbf{2}} \cdot \textbf{t} $$ Combining these formulas directly gives:
$$ \textbf{T} = \frac{\textbf{(n + m)} \cdot \textbf{(n – m + k)}}{\textbf{2k}} $$
Example:
Let us calculate the total number of terms in the arithmetic sequence: $27, 29, 31, 33, \dots, 125$.
$$\text{Number of Terms} = \frac{125 – 27}{2} + 1 = 49 + 1 = 50 \text{ terms.}$$
Example:
Let us find the total value of the sum: $T = 44 + 48 + 52 + \dots + 200$.
Since every single term in this sequence is a multiple of 4, we can factor out a 4 from the expression:
$$ T = 4 \cdot (11 + 12 + 13 + \dots + 50) $$ Here, we have an inner sequence where $m = 11$, $n = 50$, and $k = 1$. Substituting these values into the sum formula:
\[ T = 4 \cdot \frac{(50+11) \cdot (50-11+1)}{2 \cdot 1} = 4 \cdot \frac{61 \cdot 40}{2} = 4880 \]
Standard Formula Derivations:
- Sum of the first $n$ natural numbers: $1 + 2 + 3 + 4 + \dots + n$
\[ 1+2+3+4+\dots +n = \frac{(n+1)(n-1+1)}{2 \cdot 1} = \frac{n \cdot (n+1)}{2} \]
- Sum of the first consecutive even numbers up to $2n$: $2 + 4 + 6 + 8 + \dots + 2n$
\[ 2+4+6+8+\dots +2n = \frac{(2n+2)(2n-2+2)}{2 \cdot 2} = n \cdot (n+1) \]
- Sum of the first consecutive odd numbers up to $2n-1$: $1 + 3 + 5 + 7 + \dots + 2n-1$
\[ 1+3+5+7+\dots+2n-1 = \frac{(2n-1+1)(2n-1-1+2)}{2 \cdot 2} = n^2 \]
The resulting formulas for these frequently encountered arithmetic sequences are summarized below:
\[ \begin{aligned} &1 + 2 + 3 + 4 + \cdots + n = \frac{n(n + 1)}{2} \\ &2 + 4 + 6 + 8 + \cdots + 2n = n(n + 1) \\ &1 + 3 + 5 + 7 + \cdots + (2n – 1) = n^2 \end{aligned} \]
In these standard formulas, the term index value ($n$) is always found by setting the mathematical general term representation ($n$, $2n$, or $2n-1$) equal to the actual last term of the series.
For example, if $1+3+5+7+\dots+119=T$, let us find the value of $T$. We find the value of $n$ by equating the final terms:
$$119 = 2n – 1 \implies 2n = 120 \implies n = 60$$ Therefore, the sum $T$ is computed directly as:
$$T = 60^2 = 3600$$
- Given that $26+28+30+32+\dots+2n = 900$, let us determine the number of terms in this expression.
If we add the missing initial terms $2 + 4 + 6 + \dots + 24$ to both sides of the equation, we can use our standard identity:
\[ 2 + 4 + 6 + \dots + 24 + 26 + 28 + \dots + 2n = (12 \cdot 13) + 900 \]
\[ n \cdot (n+1) = 156 + 900 = 1056 \]
Since $32 \cdot 33 = 1056$, we find $n = 32$. Now, the number of terms present in our original sequence $26 + 28 + 30 + \dots + 2(32)$ is calculated as:
$$\text{Number of Terms} = \frac{64 – 26}{2} + 1 = 19 + 1 = 20 \text{ terms.}$$
- Let us find the total value of the sum: $7 + 11 + 15 + 19 + \dots + 83$.
$$ 7 + 11 + 15 + 19 + \dots + 83 = \frac{(83+7) \cdot (83-7+4)}{(2 \cdot 4)} = \frac{90 \cdot 80}{8} = 900 $$
QUESTION 8
If the sum of even natural numbers between 9 and 37 is denoted by $a$, the sum of even natural numbers between 39 and 75 is denoted by $b$, and the sum of all even natural numbers between 1 and 75 is denoted by $x$, what is the expression for $x$ in terms of $a$ and $b$?
$$ \textbf{A) } 20 + a + b \quad \textbf{B) } 58 + a + b \quad \textbf{C) } a + b – 20 \quad \textbf{D) } a + b – 58 \quad \textbf{E) } a + b – 38$$
Solution:
Let us write out the full expansion for the total sum $x$:
$$x = 2 + 4 + 6 + 8 + 10 + 12 + \dots + 36 + 38 + 40 + 42 + \dots + 74$$ Notice that we can isolate sub-sequences within this sum:
The middle segment $10 + 12 + \dots + 36$ corresponds exactly to $a$.
The final segment $40 + 42 + \dots + 74$ corresponds exactly to $b$.
Substituting these defined groupings back into the expression for $x$ leaves us with the remaining standalone values:
$$x = (2 + 4 + 6 + 8) + a + 38 + b$$
$$x = 20 + a + 38 + b \implies x = 58 + a + b$$
$\textbf{Correct Answer: B}$