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 is [2025]

Question

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 $n (A \cup B)$ is [2025]

Smart Achievers · Accepted Answer

(2)

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

$2025^{th} term of set A = 1 + 2024 \times 5 = 10121$

$2025^{th} term of set B = 9 + 2024 \times 7 = 14177$

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

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

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

$\Rightarrow$ n – 1 < 288.7

$\Rightarrow$ n < 289.7

$\Rightarrow$ n = 289 [ $∵$ n is natural number]

$\Rightarrow n (A \cup B) = n (A) + n (B) - n (A \cap B)$

= 2025 + 2025 – 289 = 3761.

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 $n (A \cup B)$ is [2025]

1 4027

2 3761

3 4003

4 3814

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