Question

Let X be the set consisting of the first 2018 terms of the arithmetic progression 1, 6, 11, _______, and Y be the set consisting of the first 2018 terms of the arithmetic 9, 16, 23, __________. Then, the number of elements in the set X$ \cup $Y is _____?

Answer 1

Hint: Before attempting this question one must have prior knowledge of sets and then remember the basic formula of sets and first find the 2018 terms in set X and Y, using this instruction find the numbers of elements in X$ \cup $Y.

Complete step-by-step solution -
According to the given information we have set X 1, 6, 11, ______, which have total 2018 terms
So in set X 1 is first term i.e. a1 =1 and a2 = 6
To find the common difference between the terms of set X we will use the formula of common difference $d = {a_2} - {a_1}$ here d is the common difference between two consecutive terms ${a_2}$ and ${a_1}$
Substituting the given values in the above formula
We get d = 6 – 1 = 5
So the common difference between the consecutive terms in the set X is 5
Since we know that set X have total 2018 terms so the last term of the set X will be
So any term of any series is given by \[{a_n} = {a_1} + \left( {n-1} \right)d\] here n is the number of the term and a1 is the first term of the set and d is the common difference between the consecutive terms in the set
Substituting the given values in the above formula
We get \[{a_{2018}} = 1 + \left( {2018-1} \right)5\]
$ \Rightarrow $${a_{2018}} = 1 + 2017 \times 5$
$ \Rightarrow $${a_{2018}}$ = 10086
So the set X will be 1, 6, 11, _______, 10086.
For set Y i.e. 9, 16, 23, _______ let b be the terms in the set Y and m be the total numbers of terms in set Y then b1 = 9 and b2 = 16 so the common difference be K between the consecutive terms of set Y will be given by \[{b_n} = {b_1} + \left( {m-1} \right)k\]
Substituting the given values in the above formula we get
k = 16 – 9 = 7
Since the set have total 2018 terms so the last term will be
\[{b_{2018}} = 9 + \left( {2018-1} \right)7\]
$ \Rightarrow $${b_{2018}} = 9 + 2017 \times 7$
$ \Rightarrow $${b_{2018}}$ = 14128
So the set Y is 9, 16, 23, _______, 14128
So the common series X $ \cap $ Y can be find as the common terms in X and Y series
Since we know that the n term in set X is given by \[{a_n} = {a_1} + \left( {n-1} \right)d\] and the n term in set Y will be given by \[{b_n} = {b_1} + \left( {m-1} \right)k\] so the condition for common terms in X and Y series will be given as
\[{b_1} + \left( {m-1} \right)k = {a_1} + \left( {n-1} \right)d\]
After using the above condition we got first common term in the set X and Y is 16 and second is 51
So X $ \cap $ Y = 16, 51, ______ where 16 is the first term and 51 is the second term of the set X $ \cap $ Y and common difference will be o = 51 – 16 = 35
Since we know that the set X $ \cap $ Y will have common terms value will be equal to or less than 10086 i.e. ${c_n} \leqslant 10086$
${c_1} + \left( {n - 1} \right)o \leqslant 10086$ Here $C_1$ is the first term of set X $ \cap $ Y
Substituting the given values in above equation
$16 + \left( {n - 1} \right)35 \leqslant 10086$
$ \Rightarrow $ $\left( {n - 1} \right) \leqslant 287$
n = 288
So set X $ \cap $ Y have 288 common terms
We have to find the numbers of element in set X$ \cup $Y which is given by
\[n\left( {X \cup Y} \right) = n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right)\]
Substituting the given values in the above equation
We get \[n\left( {X \cup Y} \right) = 2018 + 2018 - 288\]
$ \Rightarrow $\[n\left( {X \cup Y} \right) = 3748\]
Hence 3748 are the numbers of elements present in set X$ \cup $Y.

Note: In the above solution we used the concept of sets which can be explained as the collection of numbers which are so well defined, numbers in any set are named as “element”. There are some properties of sets such as when two sets have the same numbers of elements and when every element of one set is the same as the element of the second set.