(a) The fourth term of an A.P is 37 and 6th term is 12 more than the fourth term . Find the first and seventh terms. (b) If \(P = {1, 2, 3, 4}\) and \(Q = {...
Question 1 Report
(a) The fourth term of an A.P is 37 and 6th term is 12 more than the fourth term . Find the first and seventh terms.
(b) If \(P = {1, 2, 3, 4}\) and \(Q = {3, 5, 6}\), find (i) \(P \cap Q\) ; (ii) \(P \cup Q\) ; (iii) \((P \cap Q) \cup Q\) ; (iv) \((P \cap Q) \cup P\).
Answer Details
(a) Arithmetic Progression
Let the first term be \(a\) and common difference \(d\). The \(n\)th term is \(a+(n-1)d\).
Fourth term: \(a + 3d = 37\).
The 6th term is 12 more than the 4th term: \((a+5d) = 37 + 12 = 49\), so \(a + 5d = 49\).
Subtract the equations:
\[(a+5d)-(a+3d) = 49-37 \;\Rightarrow\; 2d = 12 \;\Rightarrow\; d = 6\]
Then \(a = 37 - 3(6) = 19\).
First term \(= 19\). Seventh term:
\[a + 6d = 19 + 6(6) = 55\]
Seventh term \(= 55\).
(b) Sets with \(P=\{1,2,3,4\}\) and \(Q=\{3,5,6\}\).
(i) \(P\cap Q = \{3\}\)
(ii) \(P\cup Q = \{1,2,3,4,5,6\}\)
(iii) \((P\cap Q)\cup Q = \{3\}\cup\{3,5,6\} = \{3,5,6\}\)
(iv) \((P\cap Q)\cup P = \{3\}\cup\{1,2,3,4\} = \{1,2,3,4\}\)
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