JEE Mathematics 2025 Question with Solution

Given: The A.P. has $2k$ terms. Sum of odd-position terms is $40$, sum of even-position terms is $55$, and the last term exceeds the first term by $27$.

Find: The value of $k$.

Let the first term be $a$ and common difference be $d$.

The last term is

$$a + (2k-1)d$$

Since it exceeds the first term by $27$,

$$a + (2k-1)d = a + 27$$

So,

$$(2k-1)d = 27$$

The odd terms are

$$a, a+2d, a+4d, \ldots$$

These form an A.P. with $k$ terms, so their sum is

$$S_{\text{odd}} = \frac{k}{2}\left[2a + (k-1)2d\right] = 40$$

Hence,

$$k\left(a + (k-1)d\right) = 40$$

The even terms are

$$a+d, a+3d, a+5d, \ldots$$

These also form an A.P. with $k$ terms, so their sum is

$$S_{\text{even}} = \frac{k}{2}\left[2(a+d) + (k-1)2d\right] = 55$$

Thus,

$$k(a+kd) = 55$$

Now subtract the odd-term equation from the even-term equation:

$$k(a+kd) - k\left(a + (k-1)d\right) = 55 - 40$$ $$kd = 15$$

Using

$$(2k-1)d = 27$$

and

$$kd = 15$$

we get

$$d = \frac{15}{k}$$

Substitute into the first equation:

$$(2k-1)\frac{15}{k} = 27$$ $$15(2k-1) = 27k$$ $$30k - 15 = 27k$$ $$3k = 15$$ $$k = 5$$

Therefore, the correct option is C and the value of $k$ is $5$.

The solution labels option D, but the working clearly gives $k = 5$, which matches option C.