Question

Let $A = \{1, 6, 11, 16, \ldots\}$ and $B = \{9, 16, 23, 30, \ldots\}$ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n(A \cup B)$ is

• A. 3814
• B. 4027
• C. 3761
• D. 4003

Hint:

The number of terms in the union of two sets can be found using the formula $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

Answer 1

The Correct Option is C

1. Identify the sets $A$ and $B$: - $A = \{1, 6, 11, 16, \ldots\}$ - $B = \{9, 16, 23, 30, \ldots\}$
2. Find the general terms for $A$ and $B$: - For set $A$: $T_n = 1 + (n-1) \cdot 5 = 5n - 4$ - For set $B$: $T_n = 9 + (n-1) \cdot 7 = 7n + 2$
3. Determine the intersection $A \cap B$: - Solve $5n - 4 = 7m + 2$ for $n$ and $m$: \[ 5n - 7m = 6 \] - The common terms are $16, 51, 86, \ldots$
4. Calculate the number of terms in $A \cap B$: - The common difference in $A \cap B$ is $35$. - Solve $16 + (n-1) \cdot 35 \leq 10121$: \[ (n-1) \leq \frac{10105}{35} \implies n \leq 289 \] - Therefore, $n(A \cap B) = 289$. 5. Calculate $n(A \cup B)$: \[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \] \[ n(A \cup B) = 2025 + 2025 - 289 = 3761 \] Therefore, the correct answer is (3) 3761.