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$
Step-by-Step Solution
Key Concept: Find common terms of two APs using congruences, then count among the first$2025$terms.
$A = {1$, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61,$(3)$66, 71, 76, 81, 86, 91,$\ldots$$\ldots}$B = {9$, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, 93, 100,$$\ldots$$\ldots} A$$\cap$B = {16$, 51, 86,$$\ldots$$\ldots} For set$'A' $\Rightarrow$$T2025 = 1 +$(2025 - 1)$$(5)$= 10121$′ For set ' B $\Rightarrow$$T2025 = 9 +$(2025 - 1)$$(7)$= 14177$So, for (A$\cap B)$$\Rightarrow$$T_n = 16 +$(n - 1)$(35)$$\le 10121$10121 - 16$($n - 1$)$$\$le = 288.71$35 n$$\le 289.71$\Rightarrow $n = 289$∴ n(A$\cup B) = n(A) + n(B) - n(A$$\cap B) =$2025 + 2025 - 289 = 3761$4$
Correct Answer: 3