JEE Mathematics 2025 Question with Solution

Given:

• $A = \{1, 6, 11, 16, \ldots\}$ contains the first $2025$ terms of an arithmetic progression.
• $B = \{9, 16, 23, 30, \ldots\}$ contains the first $2025$ terms of another arithmetic progression.

Find: $n(A \cup B)$.

Use the counting formula

So we only need to find the number of common terms.

The general terms are

For a common term, solve

which gives

Hence the common terms form the arithmetic progression

with common difference $35$.

The last term of set $A$ is

Therefore the common terms must satisfy

So

which gives

Thus

Now apply the union formula:

Therefore, the correct option is C.

Given: the elements of $A$ satisfy $x \equiv 1 \pmod{5}$ and the elements of $B$ satisfy $x \equiv 2 \pmod{7}$.

Find: the number of elements in $A \cup B$.

First note that

The largest term of $A$ is

and the largest term of $B$ is

For $x \in A \cap B$,

Let

Then

so

Since $7 \equiv 2 \pmod{5}$,

Hence

So write

Therefore

Now count those common terms that lie in both finite sets:

Thus

Hence $s = 0,1,2,\ldots,288$, so the number of common terms is

Finally,

Therefore, the value of $n(A \cup B)$ is $3761$, so the correct option is C.