Question

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

Answer 1

Hint: In the given question these two sets \[A\] and \[B\] contain an arithmetic progression , which means they both have separate common differences in each set . In this question we will find the first term which is common to both the sets , and then find the common arithmetic progression using the formula for finding \[{n_{th}}\] term in an A.P , which is \[{a_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term and \[d = \] common difference

Complete step-by-step answer:
Given : \[A = \left\{ {3,7,11,....,407} \right\}\] and \[B = \left\{ {2,9,16,....,709} \right\}\]
From the Arithmetic progressions of set \[A\] and set \[B\] we have a term common in them which can be calculated as :
For the set \[A\] , let the term be \[{a_n}\] which can be represented using the formula for the general term .
\[{a_n} = 3 + \left( {n - 1} \right)4\]
On solving we get ,
\[{a_n} = 4n - 1\] …. Equation (i)
Similarly for the set \[B\] , let the common term be \[{a_m}\] which can be represented as
\[{a_m} = 2 + \left( {n - 1} \right)7\]
On solving we get
\[{a_m} = 7m - 5\] … equation (ii)
Now, equating the equations (i) and (ii) we get ,
\[4n - 1 = 7m - 5\]
On solving we get ,
\[4n = 7m - 4\]
Now , on the RHS the value of \[m\] is such that on solving we get the multiple of \[4\] . Therefore , on putting \[m = 4\] we get,
\[4n = 24\]
Which is a multiple of \[4\] . Therefore , in set \[B\] the \[4th\] will be the term which is common to both the sets . Therefore ,
\[{a_4} = \left( {7 \times 4} \right) - 5\]
On solving we get ,
\[{a_4} = 23\]
Therefore , \[23\] is common in both the sets .
Now , for the common difference for the series of terms which are common in both the terms will be equal to the product of common differences of both the sets . Therefore ,
Common difference for required A.P \[ = 7 \times 4\]
On solving we get ,
 \[ = 28\]
The required A.P has last term as \[407\] .
Now , using the formula for \[{n_{th}}\] term for the number of common terms , which will be \[ \leqslant 407\]. Therefore , we have ,
\[23 + \left( {n - 1} \right)28 \leqslant 407\]
On solving we get ,
\[23 + 28n - 28 \leqslant 407\]
On solving again we get ,
\[28n \leqslant 412\]
On simplifying we get ,
\[n \leqslant \dfrac{{412}}{{28}}\]
On solving we get ,
 \[n \leqslant 14.714\]
Rounding off we get ,
\[n \leqslant 14\]
Therefore , we have \[14\] terms which are common in both the sets . Hence , option (B) is the correct answer .
So, the correct answer is “Option B”.

Note: The expression \[n\left( {A \cap B} \right)\] represents the number of intersection between set \[A\] and set \[B\] , which means all the terms that are common in set \[A\] and set \[B\] . Whenever , solving for arithmetic progression sequence, remember the formula for the general term for an A.P .