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
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- 3814
- 4027
- 3761
- 4003
The Correct Option is C
Approach Solution - 1
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.
Approach Solution -2
We have two arithmetic progressions truncated to their first 2025 terms: \(A=\{1,6,11,16,\ldots\}\) and \(B=\{9,16,23,30,\ldots\}\). We need \(n(A\cup B)=|A|+|B|-|A\cap B|\).
Concept Used:
Intersection of two arithmetic progressions can be found via simultaneous congruences. Also, for finite APs, the count of common terms equals the count of terms of the resulting AP lying within both ranges.
Step-by-Step Solution:
Step 1: Identify general terms and endpoints.
\[ A:\ a_k=1+5(k-1)=5k-4,\quad k=1,\ldots,2025\Rightarrow a_{\max}=5\cdot2025-4=10121. \] \[ B:\ b_m=9+7(m-1)=7m+2,\quad m=1,\ldots,2025\Rightarrow b_{\max}=7\cdot2025+2=14177. \] Thus \(|A|=|B|=2025\).
Step 2: Solve for common terms \(x\in A\cap B\).
\[ x\equiv 1\pmod{5}\quad(\text{since }x=5k-4),\qquad x\equiv 2\pmod{7}\quad(\text{since }x=7m+2). \] Let \(x=2+7t\). Then \(2+7t\equiv 1\pmod{5}\Rightarrow 7t\equiv -1\equiv 4\pmod{5}\). Since \(7\equiv 2\pmod{5}\), we get \(2t\equiv 4\pmod{5}\Rightarrow t\equiv 2\pmod{5}\). Hence \(t=2+5s\) and \[ x=2+7(2+5s)=16+35s,\quad s\in\mathbb{Z}. \]
Step 3: Count common terms within bounds.
\[ 16+35s\le \min(10121,14177)=10121 \Rightarrow s\le \frac{10121-16}{35}=\frac{10105}{35}=288+\frac{25}{35}. \] So \(s_{\max}=288\), and with \(s_{\min}=0\), number of common terms is \[ |A\cap B|=288-0+1=289. \]
Final Computation & Result
\[ n(A\cup B)=|A|+|B|-|A\cap B|=2025+2025-289=3761. \]
Answer: \( \boxed{3761} \)
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