JEE Mathematics 2024 Question with Solution

Given: Two arithmetic progressions are $4, 9, 14, 19, \ldots$ up to the $25$th term, and $3, 6, 9, 12, \ldots$ up to the $37$th term.

Find: The number of common terms in these two sequences.

For the first A.P., the first term is $a_1 = 4$ and common difference is $d_1 = 5$. Its $n$-th term is

So the $25$th term is

For the second A.P., the first term is $a_2 = 3$ and common difference is $d_2 = 3$. Its $n$-th term is

So the $37$th term is

The first common term is $9$. The common difference of common terms is the least common multiple of $5$ and $3$:

Hence the common terms form the A.P.

Now all common terms must lie within both given ranges. Since the second sequence goes only up to $111$, the common terms not exceeding $111$ are

There are $7$ such terms.

Therefore, the number of common terms is $7$. The correct option is C.

Given: Common terms form an arithmetic progression starting from $9$ with common difference $15$.

Find: How many such terms are within the allowed range.

Let the common terms be

So the $k$-th common term is

This must be at most $111$:

Thus,

Hence the total number of common terms is $7$. This agrees with the listed common terms and the correct option is C.