How do you find $4$ consecutive odd integers with a sum of $18\;$ ?
Answer
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Hint: Take the base number as $n$, the next consecutive odd number after $n$, which would be $(n + 2)$ and the next consecutive as $(n + 4)$ and $(n + 6)$. Write an equation adding all these terms resulting in the sum given in the question as $18\;$. Evaluate to find the value of $n$ and then substitute the value of $n$ in other terms to find the consecutive odd integers.
Complete step-by-step answer:
Firstly, To find the $4$ consecutive odd integers we take $n$ as a base number.
The next consecutive number after $n$ would be $(n + 2)$ since the immediate odd integer after ${n^{th}}$ integer occurs at an increment of $2$.
Therefore, the assumed $4$ consecutive odd integers are,
$ \Rightarrow $$n,(n + 2),(n + 4),(n + 6)$
Since the question is to find the sum of the $4$ consecutive odd integers,
Write an equation adding all the terms and equating it to the given sum $18\;$
$ \Rightarrow n + (n + 2) + (n + 4) + (n + 6) = 18$
Add all the $n$terms and then the constants.
$ \Rightarrow 4n + 12 = 18$
Subtract $12\;$ from both sides,
$ \Rightarrow 4n + 12 - 12 = 18 - 12$
$ \Rightarrow 4n = 6$
Divide with$4$on both sides.
$ \Rightarrow n = \dfrac{6}{4}$
On further simplification we get,
$ \Rightarrow n = \dfrac{3}{2}$
Now, substitute these terms to get the $4$ consecutive odd integers.
$ \Rightarrow n = \dfrac{3}{2},(n + 2 = \dfrac{3}{2} + 2),(n + 4 = \dfrac{3}{2} + 4),(n + 6 = \dfrac{3}{2} + 6)$
Evaluate further,
$ \Rightarrow n = \dfrac{3}{2},(n + 2 = \dfrac{{3 + 4}}{2}),(n + 4 = \dfrac{{3 + 8}}{2}),(n + 6 = \dfrac{{3 + 12}}{2})$
$ \Rightarrow n = \dfrac{3}{2},(n + 2 = \dfrac{7}{2}),(n + 4 = \dfrac{{11}}{2}),(n + 6 = \dfrac{{15}}{2})$
$\therefore $The $4$ consecutive odd integers are,$\dfrac{3}{2},\dfrac{7}{2},\dfrac{{11}}{2},\dfrac{{15}}{2}$
Additional information: Consecutive numbers are the numbers that follow each other in order. If it is consecutive odd numbers, then it means odd numbers which follow each other. If it is consecutive even numbers, then it means even numbers which occur one after another.
Note:
For any odd or even integer $n$, the next consecutive odd or even integers will occur at positions $(n + 2),(n + 4),(n + 6),(n + 8)...$ and so on. There might be some calculation errors in the result, the only way to cross-check it is to place it back in the equation and check if the sum of all the terms is equal to $18\;$
Complete step-by-step answer:
Firstly, To find the $4$ consecutive odd integers we take $n$ as a base number.
The next consecutive number after $n$ would be $(n + 2)$ since the immediate odd integer after ${n^{th}}$ integer occurs at an increment of $2$.
Therefore, the assumed $4$ consecutive odd integers are,
$ \Rightarrow $$n,(n + 2),(n + 4),(n + 6)$
Since the question is to find the sum of the $4$ consecutive odd integers,
Write an equation adding all the terms and equating it to the given sum $18\;$
$ \Rightarrow n + (n + 2) + (n + 4) + (n + 6) = 18$
Add all the $n$terms and then the constants.
$ \Rightarrow 4n + 12 = 18$
Subtract $12\;$ from both sides,
$ \Rightarrow 4n + 12 - 12 = 18 - 12$
$ \Rightarrow 4n = 6$
Divide with$4$on both sides.
$ \Rightarrow n = \dfrac{6}{4}$
On further simplification we get,
$ \Rightarrow n = \dfrac{3}{2}$
Now, substitute these terms to get the $4$ consecutive odd integers.
$ \Rightarrow n = \dfrac{3}{2},(n + 2 = \dfrac{3}{2} + 2),(n + 4 = \dfrac{3}{2} + 4),(n + 6 = \dfrac{3}{2} + 6)$
Evaluate further,
$ \Rightarrow n = \dfrac{3}{2},(n + 2 = \dfrac{{3 + 4}}{2}),(n + 4 = \dfrac{{3 + 8}}{2}),(n + 6 = \dfrac{{3 + 12}}{2})$
$ \Rightarrow n = \dfrac{3}{2},(n + 2 = \dfrac{7}{2}),(n + 4 = \dfrac{{11}}{2}),(n + 6 = \dfrac{{15}}{2})$
$\therefore $The $4$ consecutive odd integers are,$\dfrac{3}{2},\dfrac{7}{2},\dfrac{{11}}{2},\dfrac{{15}}{2}$
Additional information: Consecutive numbers are the numbers that follow each other in order. If it is consecutive odd numbers, then it means odd numbers which follow each other. If it is consecutive even numbers, then it means even numbers which occur one after another.
Note:
For any odd or even integer $n$, the next consecutive odd or even integers will occur at positions $(n + 2),(n + 4),(n + 6),(n + 8)...$ and so on. There might be some calculation errors in the result, the only way to cross-check it is to place it back in the equation and check if the sum of all the terms is equal to $18\;$
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