Question

If $A=\{4n+2|n$is a natural number$\}$ and $B=\{3n|n$is a natural number$\}$then what is $A\cap B$ equal to
A. $\{12{{n}^{2}}+6n|n$is a natural number$\}$
B. \[\{24n-12|n\]is a natural number$\}$
C. \[\{60n+30|n\] is a natural number$\}$
D. \[\{12n-6|n\] is a natural number$\}$

Answer 1

Hint: To solve this question we will first form a set of $A$ and $B$ by substituting some natural values of $n$ in $A=4n+2$ and $B=3n$. Then we will find $A\cap B$ by taking common elements from set $A$ and $B$. We will then see the pattern in the elements of $A\cap B$ and form general relation of it in terms of $n$.

Complete step by step solution: We are given $A=\{4n+2|n$is a natural number$\}$ and $B=\{3n|n$is a natural number$\}$ and we have to determine the value of $A\cap B$.
As given that $n$is a natural number we will put some values of $n$ and derive some values of $A$ and $B$.
Substituting $n=1,2,3,4,5,6.....$ in $A$ and $B$.
$\begin{align}
  & A=\{6,10,14,18,22,26,30....\} \\
 & B=\{3,6,9,12,15,18,21....\} \\
\end{align}$
We know that $A\cap B$ which is read as “ $A$ intersection $B$” can be defined as the set of all the elements which is common in both set $A$ and in set $B$. So,
$A\cap B=\{6,18,30,..\}$
From above set of $A\cap B$ we can see that there is a series forming with an increment of $12$ in each of the next number. That is,
$\begin{align}
  & 6 \\
 & 6+12=18 \\
 & 18+12=30 \\
\end{align}$
So in terms of $n$ we can write $A\cap B=\{6,18,30,..\}$ as,
$\begin{align}
  & A\cap B=\{6+(n-1)12\} \\
 & =6+12n-12 \\
 & =12n-6
\end{align}$
So the answer will be \[\{12n-6|n\] is a natural number$\}$.
If $A=\{4n+2|n$is a natural number$\}$ and $B=\{3n|n$is a natural number$\}$then $A\cap B$ is equal to \[\{12n-6|n\] is a natural number$\}$


Option ‘D’ is correct

Note: We can check by substituting natural values of $n$ in \[12n-6\] to see if we will get the same series as $A\cap B=\{6,18,30,..\}$ or not.