What are three consecutive integers whose sum is $96$?
Answer
552.9k+ views
Hint:Consecutive integers are those numbers that follow each other. They follow a sequence or order. Three consecutive integers means three numbers that follow each other like $2,3,4$. So, if we add these up we get $9$ as a sum. So, in question we need to find three consecutive integers that follow each other and when we add these numbers we will get a sum of $96$.
Complete step by step answer:
In the question we need to find three consecutive integers whose sum is $96$ we will use algebra to find these numbers. Let’s assume the first integer to be $x$. Since the integers are consecutive. Second integer would be $x + 1$. Third integer would be $x + 2$. When we add these consecutive integers we have a sum of $96$. Therefore, we can write it as
$(x) + (x + 1) + (x + 2) = 96$
Now to solve for $x$, we first add the integers together and $x$ variables together.
So, the above equation becomes
$ \Rightarrow $$3x + 3 = 96$
Shifting $3$ to the right side of the equation. We get,
$ \Rightarrow 3x = 96 - 3$
$ \Rightarrow 3x = 93$
$ \Rightarrow x = \dfrac{{93}}{3}$
$ \Rightarrow x = 31$
So, first integer is $x = 31$
Second integer $ = x + 2 = 31 + 1 = 32$
Third integer $ = x + 2 = 31 + 2 = 33$
Therefore, three consecutive integers that add up to $96$ are $31,\,32$ and $33$.
Note:Consecutive integers are integers that follow each other in a fixed sequence, each number being $1$ more than the previous number. Consecutive numbers are of three types- Normal consecutive numbers, even consecutive numbers and odd consecutive numbers. In Normal consecutive numbers each number is $1$ more than the previous numbers and in odd and even consecutive numbers the numbers follow each other by the difference of $2$. We can represent the normal consecutive integer by the formula $n + 1$ where $n = 0,1,2,3 \ldots $ similarly we can represent odd and even consecutive integers by the formula $n + 2$ where $n = 0,1,3 \ldots $.
Complete step by step answer:
In the question we need to find three consecutive integers whose sum is $96$ we will use algebra to find these numbers. Let’s assume the first integer to be $x$. Since the integers are consecutive. Second integer would be $x + 1$. Third integer would be $x + 2$. When we add these consecutive integers we have a sum of $96$. Therefore, we can write it as
$(x) + (x + 1) + (x + 2) = 96$
Now to solve for $x$, we first add the integers together and $x$ variables together.
So, the above equation becomes
$ \Rightarrow $$3x + 3 = 96$
Shifting $3$ to the right side of the equation. We get,
$ \Rightarrow 3x = 96 - 3$
$ \Rightarrow 3x = 93$
$ \Rightarrow x = \dfrac{{93}}{3}$
$ \Rightarrow x = 31$
So, first integer is $x = 31$
Second integer $ = x + 2 = 31 + 1 = 32$
Third integer $ = x + 2 = 31 + 2 = 33$
Therefore, three consecutive integers that add up to $96$ are $31,\,32$ and $33$.
Note:Consecutive integers are integers that follow each other in a fixed sequence, each number being $1$ more than the previous number. Consecutive numbers are of three types- Normal consecutive numbers, even consecutive numbers and odd consecutive numbers. In Normal consecutive numbers each number is $1$ more than the previous numbers and in odd and even consecutive numbers the numbers follow each other by the difference of $2$. We can represent the normal consecutive integer by the formula $n + 1$ where $n = 0,1,2,3 \ldots $ similarly we can represent odd and even consecutive integers by the formula $n + 2$ where $n = 0,1,3 \ldots $.
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