Here are tips for identifying set theory problems and the solutions to the questions: Tips to Identify Set Theory Problems Set theory problems deal with collections of objects, called sets, and the relationships and operations between them. Here are some common indicators: Symbols: Look for specific set notation symbols: $\cup$ (Union): "or", "at least one", "all elements in either set". $\cap$ (Intersection): "and", "both", "common elements". $'$ or $^c$ (Complement): "not in", "elements outside the set". $\setminus$ or $-$ (Difference): "in A but not in B". $\subseteq$ (Subset): "is contained in". $\in$ (Element of): "is a member of". $\emptyset$ (Empty set): A set with no elements. $U$ (Universal set): The set of all possible elements under consideration. Keywords: Phrases like "equivalence of sets," "simplify the set expression," "prove the identity," "number of elements," "survey results," "groups of people," "at least one," "neither," "only." Context: Problems involving grouping, categorization, or relationships between different collections of items (e.g., students taking different subjects, people with different hobbies). Venn Diagrams: If the problem mentions or can be easily visualized with overlapping circles, it's likely a set theory problem. Solutions to the Questions 4. The equivalence of $A \cap (B \cup C)$ is: a) $(A \cup B') \cup (A \cup C)$ b) $(A \cap B) \cup C$ c) $(A \cap B) \cup (A \cap C)$ d) $(A \cup B) \cap (A \cup C)$ This question asks for an equivalent expression using the Distributive Law of set operations. The Distributive Law states that for any sets A, B, and C: $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$ This law is analogous to distributing multiplication over addition in algebra, e.g., $a \times (b + c) = (a \times b) + (a \times c)$. Comparing the given expression with the options, option (c) directly matches the Distributive Law. The correct option is $\boxed{\text{c) } (A \cap B) \cup (A \cap C)}$. 5. For an AP, $a=25, d=-3$ find the value of $n$ if $S_n = 112$. Step 1: Identify the given values for the Arithmetic Progression (AP). First term: $a = 25$ Common difference: $d = -3$ Sum of the first $n$ terms: $S_n = 112$ Step 2: Use the formula for the sum of the first $n$ terms of an AP. The formula is: $$S_n = \frac{n}{2} [2a + (n-1)d]$$ Step 3: Substitute the given values into the formula. $$112 = \frac{n}{2} [2(25) + (n-1)(-3)]$$ Step 4: Simplify the equation. $$112 = \frac{n}{2} [50 - 3(n-1)]$$ $$112 = \frac{n}{2} [50 - 3n + 3]$$ $$112 = \frac{n}{2} [53 - 3n]$$ Multiply both sides by 2 to eliminate the fraction: $$2 \times 112 = n(53 - 3n)$$ $$224 = 53n - 3n^2$$ Step 5: Rearrange the equation into a standard quadratic form ($Ax^2 + Bx + C = 0$). $$3n^2 - 53n + 224 = 0$$ Step 6: Solve the quadratic equation for $n$ using the quadratic formula $n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$. Here, $A=3$, $B=-53$, $C=224$. $$n = \frac{-(-53) \pm \sqrt{(-53)^2 - 4(3)(224)}}{2(3)}$$ $$n = \frac{53 \pm \sqrt{2809 - 2688}}{6}$$ $$n = \frac{53 \pm \sqrt{121}}{6}$$ $$n = \frac{53 \pm 11}{6}$$ This gives two possible values for $n$: $$n_1 = \frac{53 + 11}{6} = \frac{64}{6} = \frac{32}{3}$$ $$n_2 = \frac{53 - 11}{6} = \frac{42}{6} = 7$$ Since the number of terms ($n$) must be a positive integer, $n = \frac{32}{3}$ is not a valid solution. Therefore, the value of $n$ is 7. The value of $n$ is $\boxed{7}$.