Let X be the set consisting of the first 2018 terms of the arithmetic progression 1,6,11,… , and Y be the set consisting of the first 2018 terms of the arithmetic progression 9,16,23,… . Then, the number of elements in the set X∪Y is___.

JEE Advanced · Mathematics · 5. Sequences & Series

Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1,6, 11, \dots,$ and $Y$ be the set consisting of the first $2018$ terms of the arithmetic progression $9,16,23, \dots$ . Then, the number of elements in the set $X \cup Y$ is___.

A 3750
B 3748
C 1245
D 6487

Verified Solution

Answer & Solution

Correct Answer

(B) 3748

Step-by-step Solution

Detailed explanation

X : 1, 6, 11, \dots, 10086

Y : 9, 16, 23, \dots ., 14128

X \cap Y : 16, 51, 86, \dots . .

Let

m = n (X \cap Y)

∴ 16 + (m - 1) \times 35 \leq 10086

\Rightarrow m \leq 288.71

\Rightarrow m = 288

∴ n (X \cup Y) = n (X) + n (Y) - n (X \cap Y)

= 2018 + 2018 - 288 = 3748

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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