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Solution


Q.

Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A∪B) is          [2025]
1 4027  
2 3761  
3 4003  
4 3814  

Ans.

(2)

Given A = {1, 6, 11, 16, ...}, B = {9, 16, 23, 30, ...} are two A.P. and n(A) = 2025, n(B) = 2025

2025th term of set A=1+2024×5=10121

2025th term of set B=9+2024×7=14177

⇒ A = {1, 6, 11, 16, ..., 10121} and B = {9, 16, 23, ...,14177}

Now A∩B = 16, 51, 86, .... which is an A.P. with a = 16, d = 35

then number of terms in (A∩B) = 16 + (n – 1) 35 < 10121

⇒ n – 1 < 288.7

⇒ n < 289.7

⇒ n = 289              [∵ n is natural number]

⇒ n(A∪B)=n(A)+n(B)–n(A∩B)

                          = 2025 + 2025 – 289 = 3761.