Question

 If a set \[A = \left\{ {3,7,11,...,407} \right\}\] and a set \[B = \left\{ {2,9,16,...,709} \right\}\]. Then find the value of \[n\left( {A \cap B} \right)\].
A. \[13\]
B. \[14\]
C. \[15\]
D. \[16\]

Answer 1

Hint:
First, find the common difference for the both sets \[A\] and \[B\]. Then, use the common differences of both sets to calculate the first term and the common difference of the common series. After that, use the formula of the \[{n^{th}}\] term of an arithmetic progression and calculate the number of elements of the common series. In the end, simplify the inequality equation to get the required answer.


Formula Used:
The \[{n^{th}}\] term of an arithmetic progression: \[{a_n} = a + \left( {n - 1} \right)d\]


Complete step-by-step answer:
The given sets are \[A = \left\{ {3,7,11,...,407} \right\}\] and \[B = \left\{ {2,9,16,...,709} \right\}\]
Clearly, we can see that the elements of both sets are in an arithmetic progression.
Common difference in set A = 4 and
Common difference in set B = 7

First term of the common series = \[23\]
Common difference = \[4 \times 7 = 28\]
The terms of the common series are = \[23,51,79,107,....\]

Since the last term of the common series \[\le 407\]
Let the number of elements present in the common series = \[p\]
Apply the formula of the \[{n^{th}}\] term of an arithmetic progression to calculate the number of elements of the common series.
We get,
\[23 + \left( {p - 1} \right)28 \le 407\]
\[ \Rightarrow 23 + 28p - 28 \le 407\]
\[ \Rightarrow 28p - 5 \le 407\]
\[\Rightarrow 28p \le 412\]
Divide both sides by \[28\]
\[\Rightarrow p \le 14.71\]
Since \[p\] is a natural number.
\[p = 14\]
Therefore, \[n\left( {A \cap B} \right) = p = 14\]
Hence the correct option is B.



Note:
Students often make a common mistake that is they are using the formula for \[{n^{th}}\] term of an arithmetic progression as \[{a_n} = a + nd\] which is a wrong formula. The correct formula is \[{a_n} = a + \left( {n - 1} \right)d\].