JEE Mathematics 2025 Question with Solution

Given: The A.P. has an even number of terms. The sum of terms in odd positions is $24$, the sum of terms in even positions is $30$, and the last term exceeds the first by $\frac{21}{2}$.

Find: The number of terms which are integers in the A.P.

Let the total number of terms be $2m$, first term be $a$ and common difference be $d$.

Then,

and

Subtracting the two sums,

Each bracket equals $d$, and there are $m$ such brackets. Therefore,

Also, the last term is $a_{2m}=a+(2m-1)d$. Since the last term exceeds the first term by $\frac{21}{2}$,

Using $md=6$, we get $d=\frac{6}{m}$. Substitute into the previous relation:

Hence the total number of terms is

and

Now use the sum of odd-position terms. The odd-position terms form an A.P. with first term $a$, common difference $2d$, and number of terms $4$. So,

Since $d=\frac{3}{2}$,

Therefore the A.P. is

The integer terms are $3,6,9,12$, which are $4$ in number.

Therefore, the correct option is A.

The solution contains arithmetic inconsistencies in the listed A.P., but the working leading to $a=\frac{3}{2}$, $d=\frac{3}{2}$ and total terms $8$ supports that the number of integer terms is $4$.

Given: Odd-position sum is $24$ and even-position sum is $30$.

Find: A quick relation between the number of pairs and the common difference.

Write the A.P. as

Then,

and similarly every even-position term exceeds the previous odd-position term by exactly $d$.

So when the sum of odd-position terms is subtracted from the sum of even-position terms, we obtain $m$ copies of $d$:

Given that this difference is $30-24=6$,

This is the key simplification that makes the problem direct.