Question

Let \[A = \{ n \in N:n\] is a three digit natural number \[\} \], \[B = \{ 9k + 2:k \in N\} \] and \[C = \{ 9k + I:k \in N\} \] for some \[I(0 < I \le 9)\]. If some of all the elements of set \[A \cap (B \cup C)\] is \[274 \times 400\], then what is the value of \[I\] ?

Answer 1

Hint: Find the possible elements of the individual sets \[A,B,C\] which are three-digit numbers satisfying the given conditions are found out and then the sum of the elements of \[A \cap (B \cup C)\] is calculated in terms of \[I\] to establish the equation with the given numerical sum

Formula Used: The last term of an A.P. is \[l = a + (n - 1)d\] and the sum of \[n\]-terms of an A.P. is \[{S_n} = \frac{n}{2}(a + l)\], where,
\[a = \]the first term of A.P.
\[l = \]the last term of A.P.
\[d = \]the common difference
\[n = \]the number of terms
\[{S_n} = \]the sum of \[n\]-terms

Complete step-by-step solution: We have been given three sets with the following conditions
\[A = \{ n \in N:n\] is a three digit natural number\[\} \]
\[B = \{ 9k + 2:k \in N\} \]
\[C = \{ 9k + I:k \in N\} \] for some \[I(0 < I \le 9)\]
Also we have been given the sum of the elements of the set \[A \cap (B \cup C)\] as \[274 \times 400\].
First, we have to find out the elements of \[A \cap (B \cup C)\].
Given, \[A\] is a set of three digit natural numbers only.
So, all the elements of \[A \cap (B \cup C)\] will be three digit natural numbers only.
Let, us find out the elements of the three sets which are three digit natural numbers only satisfying the given conditions.
\[A = \{ n \in N:n\] is a three digit natural number\[\} \]
So, \[A = \{ 100,101,102,103,................,999\} \]
Finding three digit numbers in the form\[9k + 2\] as given in the set \[B = \{ 9k + 2:k \in N\} \], we have
\[9k + 2 = 101\] for \[k = 11\], \[9k + 2 = 110\] for \[k = 12\], …………………………………………,\[9k + 2 = 992\] for \[k = 110\].
So, \[B = \{ 101,110,119,.................................,992\} \], which is an A.P., in which,
The first term \[ = 101 = a\]
The last term \[ = 992 = l\]
The common difference \[ = 9 = d\]
Let, the number of terms \[ = n\]
Applying the formulae, \[l = a + (n - 1)d\], we have
\[992 = 101 + (n - 1)9\]
\[ \Rightarrow (n - 1)9 = 891\]
\[ \Rightarrow (n - 1) = 99\]
\[ \Rightarrow n = 100\]
So, the number of terms in the set \[B\] is \[100\].
Now, applying the formulae \[{S_n} = \frac{n}{2}(a + l)\], we will find the sum of the elements of set \[B\].
Let, \[{S_1} = \] the sum of the elements of set \[B\].
Substituting the values, we have
\[{S_1} = \frac{{100}}{2}(101 + 992) = 50 \times 1093 = 54650\]
Given, \[C = \{ 9k + I:k \in N\} \] for some \[I(0 < I \le 9)\]
Now, we will find the first and last term of \[C\] in terms of \[I\] with given conditions.
The smallest three digit number is \[100\], which may be the first term of set \[C\], if it satisfies the given condition.
Let, \[I = 1\]
Now, \[9k + 1 = 100 \Rightarrow k = 11\], which is a natural number.
So, the first term of \[C\] in terms of \[I\] is \[99 + I\]
Let us examine the possible last term in terms of \[I\].
The largest three digit number \[ = 999\]
Considering \[I = 9\],we have \[9k + 9 = 999 \Rightarrow k = 110\], which is a natural number.
So, the last term of \[C\] in terms of \[I\] is \[990 + I\].
So, \[C = \{ 99 + I,108 + I,117 + I,.................................,990 + I\} \], which is an A.P., in which,
The first term \[ = 99 + I = a\]
The last term \[ = 990 + I = l\]
The common difference \[ = 9 = d\]
Let, the number of terms \[ = n\]
Applying the formulae, \[l = a + (n - 1)d\], we have
\[990 + I = 99 + I + (n - 1)9\]
\[ \Rightarrow 990 = 99 + (n - 1)9\]
\[ \Rightarrow (n - 1)9 = 891\]
\[ \Rightarrow (n - 1) = 99\]
\[ \Rightarrow n = 100\]
So, the number of terms in the set \[C\] is \[100\].
Let, \[{S_2} = \] the sum of the elements of set \[C\].
Now, applying the formulae \[{S_n} = \frac{n}{2}(a + l)\], we will find the sum of the elements of set \[C\].
Substituting the values, we have
\[{S_2} = \frac{{100}}{2}(99 + I + 990 + I) = 50 \times (1089 + 2I)\]
It is given that the sum of the elements of the set \[A \cap (B \cup C)\]is \[274 \times 400\].
So, \[{S_1} + {S_2} = 274 \times 400\]
\[ \Rightarrow 54650 + 50(1089 + 2I) = 109600\]
\[ \Rightarrow 50(1089 + 2I) = 54950\]
\[ \Rightarrow 1089 + 2I = 1099\]
\[ \Rightarrow I = 5\]

Hence, the value of \[I\] is equal to \[5\] .

Note:Carefully, the elements of \[A \cap (B \cup C)\] should be decided with the given conditions and then the formulae for the sum of \[n\]terms of A.P. i.e. \[{S_n} = \frac{n}{2}(a + l)\] should be applied to find the sum of the elements of .\[A \cap (B \cup C)\]. to establish the equation with the given data. The formulae is not applicable for the terms of a G.P.