Suppose that the number of terms in an A.P. is $2k, k \in \mathbb{N}$. If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then $k$ is equal to:

Step 1: Define the A.P. and its terms

Let the arithmetic progression (A.P.) be denoted by $a_1, a_2, \dots, a_{2k}$, where $a_1$ is the first term and $d$ is the common difference. The total number of terms is $2k$.

Step 2: Formulate equations from given sums

The odd terms form an A.P. with $k$ terms, first term $a_1$, and common difference $2d$. The sum of odd terms is given as 40. The even terms also form an A.P. with $k$ terms, first term $a_2 = a_1+d$, and common difference $2d$. The sum of even terms is given as 55. We use the sum formula $S_n = \frac{n}{2}(2a + (n-1)d')$.

Step 3: Subtract the sum equations

Subtracting equation (1) from equation (2) simplifies the expression significantly, allowing us to find a relationship between $k$ and $d$.

Step 4: Solve for $kd$

After subtracting the equations, most terms cancel out, leaving us with a simple equation $kd = 15$. This is a crucial relationship.

Step 5: Use the last term condition

The problem states that the last term, $a_{2k}$, exceeds the first term, $a_1$, by 27. We use the formula for the $n$ -th term of an A.P., $a_n = a_1 + (n-1)d$, to express $a_{2k}$.

Step 6: Solve for $k$

Substituting $kd=15$ from equation (3) into the simplified last term condition, we get $2(15) - d = 27$. This allows us to solve for $d$ and then for $k$.

Step 7: Calculate $d$ and $k$

From $30 - d = 27$, we find $d=3$. Then, using $kd=15$, we substitute $d=3$ to find $k=5$.

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