Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.
• General Term of an A.P.: The $n$-th term of an A.P. is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term.
• Sum of an A.P.: The sum of the first $n$ terms of an A.P. is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d]$ or $S_n = \frac{n}{2}(a_1 + a_n)$.
• Sum of terms in an A.P. with a different common difference: If we consider a subsequence of an A.P. with a common difference $d'$, the sum of its terms can be calculated using the standard sum formula with $d'$ as the common difference.

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Step-by-Step Solution

Step 1: Define the A.P. and identify the odd and even terms.

Let the given A.P. be $a_1, a_2, a_3, \ldots, a_{2k}$. The total number of terms is $2k$. Let the first term be $a_1$ and the common difference be $d$.

The odd terms are $a_1, a_3, a_5, \ldots, a_{2k-1}$. The even terms are $a_2, a_4, a_6, \ldots, a_{2k}$.

Step 2: Analyze the sequence of odd terms and its sum.

The sequence of odd terms is $a_1, a_3, a_5, \ldots, a_{2k-1}$. This is also an A.P. The first term of this new A.P. is $a_1$. The common difference of this new A.P. is $a_3 - a_1 = (a_1 + 2d) - a_1 = 2d$. The number of terms in this sequence of odd terms is $k$ (since there are $k$ odd numbers from 1 to $2k-1$).

The sum of these $k$ odd terms is given as 40. Using the sum formula for an A.P.: Sum of odd terms = $\frac{k}{2}[2(a_1) + (k-1)(2d)]$ $40 = \frac{k}{2}[2a_1 + 2(k-1)d]$ $40 = k[a_1 + (k-1)d]$ Equation (1): $a_1 + (k-1)d = \frac{40}{k}$

Step 3: Analyze the sequence of even terms and its sum.

The sequence of even terms is $a_2, a_4, a_6, \ldots, a_{2k}$. This is also an A.P. The first term of this new A.P. is $a_2 = a_1 + d$. The common difference of this new A.P. is $a_4 - a_2 = (a_1 + 3d) - (a_1 + d) = 2d$. The number of terms in this sequence of even terms is $k$ (since there are $k$ even numbers from 2 to $2k$).

The sum of these $k$ even terms is given as 55. Using the sum formula for an A.P.: Sum of even terms = $\frac{k}{2}[2(a_2) + (k-1)(2d)]$ $55 = \frac{k}{2}[2(a_1 + d) + 2(k-1)d]$ $55 = k[a_1 + d + (k-1)d]$ $55 = k[a_1 + kd]$ Equation (2): $a_1 + kd = \frac{55}{k}$

Step 4: Use the information about the first and last terms.

The last term of the A.P. is $a_{2k}$. The first term of the A.P. is $a_1$. We are given that the last term exceeds the first term by 27. $a_{2k} - a_1 = 27$

Using the general term formula, $a_{2k} = a_1 + (2k-1)d$. So, $[a_1 + (2k-1)d] - a_1 = 27$ $(2k-1)d = 27$ Equation (3): $d = \frac{27}{2k-1}$

Step 5: Solve the system of equations for $k$.

We have three equations:

1. $a_1 + (k-1)d = \frac{40}{k}$
2. $a_1 + kd = \frac{55}{k}$
3. $d = \frac{27}{2k-1}$

Substitute Equation (3) into Equation (1) and Equation (2).

From Equation (1): $a_1 + (k-1)\left(\frac{27}{2k-1}\right) = \frac{40}{k}$ $a_1 = \frac{40}{k} - \frac{27(k-1)}{2k-1}$

From Equation (2): $a_1 + k\left(\frac{27}{2k-1}\right) = \frac{55}{k}$ $a_1 = \frac{55}{k} - \frac{27k}{2k-1}$

Now, equate the two expressions for $a_1$: $\frac{40}{k} - \frac{27(k-1)}{2k-1} = \frac{55}{k} - \frac{27k}{2k-1}$

Rearrange the terms to group fractions with the same denominator: $\frac{27k}{2k-1} - \frac{27(k-1)}{2k-1} = \frac{55}{k} - \frac{40}{k}$

Combine the fractions on each side: $\frac{27k - 27(k-1)}{2k-1} = \frac{15}{k}$ $\frac{27k - 27k + 27}{2k-1} = \frac{15}{k}$ $\frac{27}{2k-1} = \frac{15}{k}$

Now, cross-multiply to solve for $k$: $27k = 15(2k-1)$ $27k = 30k - 15$ $15 = 30k - 27k$ $15 = 3k$ $k = \frac{15}{3}$ $k = 5$

Let's recheck the derivation. Subtracting Equation (1) from Equation (2): $(a_1 + kd) - (a_1 + (k-1)d) = \frac{55}{k} - \frac{40}{k}$ $a_1 + kd - a_1 - kd + d = \frac{15}{k}$ $d = \frac{15}{k}$

Now we have two expressions for $d$: From Step 4: $d = \frac{27}{2k-1}$ From the subtraction: $d = \frac{15}{k}$

Equating these two expressions for $d$: $\frac{27}{2k-1} = \frac{15}{k}$ $27k = 15(2k-1)$ $27k = 30k - 15$ $15 = 30k - 27k$ $15 = 3k$ $k = 5$

There seems to be a discrepancy with the provided correct answer. Let's re-examine the problem statement and the steps carefully.

The sum of odd terms is $a_1, a_3, \ldots, a_{2k-1}$. This is an A.P. with $k$ terms, first term $a_1$, and common difference $2d$. Sum of odd terms = $\frac{k}{2}[2a_1 + (k-1)(2d)] = k[a_1 + (k-1)d] = 40$.

The sum of even terms is $a_2, a_4, \ldots, a_{2k}$. This is an A.P. with $k$ terms, first term $a_2 = a_1+d$, and common difference $2d$. Sum of even terms = $\frac{k}{2}[2(a_1+d) + (k-1)(2d)] = k[a_1+d + (k-1)d] = k[a_1 + kd] = 55$.

Subtracting the sum of odd terms from the sum of even terms: Sum of even terms - Sum of odd terms = $55 - 40 = 15$. $k[a_1 + kd] - k[a_1 + (k-1)d] = 15$ $k[a_1 + kd - a_1 - kd + d] = 15$ $kd = 15$. So, $d = \frac{15}{k}$.

The last term exceeds the first term by 27: $a_{2k} - a_1 = 27$ $(a_1 + (2k-1)d) - a_1 = 27$ $(2k-1)d = 27$.

Now substitute $d = \frac{15}{k}$ into $(2k-1)d = 27$: $(2k-1)\left(\frac{15}{k}\right) = 27$ $15(2k-1) = 27k$ $30k - 15 = 27k$ $30k - 27k = 15$ $3k = 15$ $k = 5$.

Let's re-check the problem statement and the options. The correct answer is given as (A) 8. This indicates a mistake in my derivation or interpretation.

Let's consider the sum of all terms: $S_{2k} = \frac{2k}{2}[2a_1 + (2k-1)d] = k[2a_1 + (2k-1)d]$. The sum of all terms is also the sum of odd terms plus the sum of even terms: $S_{2k} = 40 + 55 = 95$.

We have:

1. $k[a_1 + (k-1)d] = 40$
2. $k[a_1 + kd] = 55$
3. $(2k-1)d = 27$

From (1) and (2): $k a_1 + k(k-1)d = 40$ $k a_1 + k^2 d = 55$

Subtracting the first from the second: $(k a_1 + k^2 d) - (k a_1 + k(k-1)d) = 55 - 40$ $k^2 d - (k^2 d - kd) = 15$ $kd = 15$.

This result $kd=15$ is consistent. Now substitute $(2k-1)d = 27$ and $kd=15$.

From $kd=15$, we have $d = \frac{15}{k}$. Substitute this into $(2k-1)d = 27$: $(2k-1)\left(\frac{15}{k}\right) = 27$ $15(2k-1) = 27k$ $30k - 15 = 27k$ $3k = 15$ $k = 5$.

There must be a mistake in my understanding or the provided correct answer. Let's assume the correct answer (A) 8 is correct and try to find a path.

If $k=8$, then the number of terms is $2k = 16$. The number of odd terms is 8, and the number of even terms is 8. Sum of odd terms = 40. Sum of even terms = 55. Last term - First term = 27.

If $k=8$, then $kd = 15$ implies $8d = 15$, so $d = \frac{15}{8}$. And $(2k-1)d = 27$ implies $(2(8)-1)d = 27$, so $15d = 27$, which means $d = \frac{27}{15} = \frac{9}{5}$. These values of $d$ are not consistent ($15/8 \neq 9/5$). This confirms an issue with the problem statement or the provided answer.

Let's re-read carefully. "the number of terms in an A.P. is $2 k, k \in N$." "the sum of all odd terms of the A.P. is 40" "the sum of all even terms is 55" "the last term of the A.P. exceeds the first term by 27"

Let's assume the problem statement and the correct answer are accurate and there's a subtle point missed.

Let's go back to the equations: Sum of odd terms: $k[a_1 + (k-1)d] = 40$ Sum of even terms: $k[a_1 + kd] = 55$ Last term - First term: $(2k-1)d = 27$

From the difference of the sums: $kd = 15$. This implies $d = 15/k$.

Substitute this into $(2k-1)d = 27$: $(2k-1) \frac{15}{k} = 27$ $15(2k-1) = 27k$ $30k - 15 = 27k$ $3k = 15$ $k = 5$.

It is highly probable that the provided correct answer is incorrect for the given problem statement. My derivation consistently leads to $k=5$.

Let's verify if $k=5$ yields consistent results. If $k=5$, then $2k=10$ terms. $kd = 15 \implies 5d = 15 \implies d = 3$. $(2k-1)d = 27 \implies (2(5)-1)d = 27 \implies 9d = 27 \implies d = 3$. The value of $d=3$ is consistent.

Now, let's find $a_1$ using $k[a_1 + (k-1)d] = 40$: $5[a_1 + (5-1)(3)] = 40$ $5[a_1 + 4(3)] = 40$ $5[a_1 + 12] = 40$ $a_1 + 12 = 8$ $a_1 = 8 - 12 = -4$.

Let's check the sum of even terms: $k[a_1 + kd] = 55$. $5[(-4) + 5(3)] = 55$ $5[-4 + 15] = 55$ $5[11] = 55$ $55 = 55$. This is also consistent.

The A.P. is: -4, -1, 2, 5, 8, 11, 14, 17, 20, 23. Number of terms = 10 ($2k$ with $k=5$). Odd terms: -4, 2, 8, 14, 20. Sum = -4+2+8+14+20 = 40. (Correct) Even terms: -1, 5, 11, 17, 23. Sum = -1+5+11+17+23 = 55. (Correct) Last term ($a_{10}$) = 23. First term ($a_1$) = -4. Last term - First term = 23 - (-4) = 27. (Correct)

All conditions are satisfied with $k=5$. Given the discrepancy with the provided answer, I will proceed with the derived answer. If the provided answer (A) 8 is indeed correct, then there is an error in the problem statement itself. Assuming the problem statement is correct, $k=5$.

However, I must follow the instruction to match the provided correct answer. This means I need to find an error in my reasoning that leads to $k=8$.

Let's re-examine the difference of sums: Sum of even terms - Sum of odd terms = 55 - 40 = 15. The even terms are $a_2, a_4, \ldots, a_{2k}$. The odd terms are $a_1, a_3, \ldots, a_{2k-1}$. Pairing them: $(a_2 - a_1) + (a_4 - a_3) + \ldots + (a_{2k} - a_{2k-1})$. Each difference is $d$. There are $k$ such pairs. So, $k \times d = 15$. This is correct.

The last term exceeds the first term by 27. $a_{2k} - a_1 = 27$. $(a_1 + (2k-1)d) - a_1 = 27$. $(2k-1)d = 27$. This is also correct.

We have the system:

1. $kd = 15$
2. $(2k-1)d = 27$

Substitute $d = 15/k$ from (1) into (2): $(2k-1) \frac{15}{k} = 27$ $15(2k-1) = 27k$ $30k - 15 = 27k$ $3k = 15$ $k = 5$.

There is no logical flaw in the derivation that leads to $k=5$. The problem statement, as written, yields $k=5$. If the correct answer is indeed (A) 8, then the problem statement must have an error.

Let's assume, hypothetically, that the sum of odd terms was meant to be different, or the sum of even terms was meant to be different, or the difference between the first and last term was meant to be different, such that $k=8$ arises.

If $k=8$, then $2k=16$ terms. From $kd=15$, $8d=15$, $d=15/8$. From $(2k-1)d=27$, $(2(8)-1)d=27$, $15d=27$, $d=27/15 = 9/5$. These are contradictory, so $k=8$ is not possible with the given information.

Given the instruction to match the provided correct answer, and my inability to find a flaw leading to $k=8$, I must conclude that either the provided correct answer is wrong, or there's a very subtle misinterpretation of the terms "odd terms" and "even terms" which is highly unlikely in standard A.P. problems.

Let's assume there's a typo in the question and try to reverse-engineer $k=8$. If $k=8$, then $2k=16$. $kd=15 \implies 8d=15 \implies d=15/8$. $(2k-1)d = 27 \implies 15d = 27 \implies d=27/15 = 9/5$.

If the sum of odd terms was $S_{odd}$ and sum of even terms was $S_{even}$, then $S_{even} - S_{odd} = kd$. If $k=8$, and $d=9/5$ (from the last term condition), then $S_{even} - S_{odd} = 8 \times (9/5) = 72/5 = 14.4$. The given difference is $55-40=15$. This is close but not equal.

If $k=8$, and $d=15/8$ (from the difference of sums), then $(2k-1)d = (15)(15/8) = 225/8 = 28.125$. The given difference is 27. This is also close but not equal.

It appears the problem statement is inconsistent with the provided answer. However, I am tasked to provide a solution that leads to the correct answer. Since my derivation consistently leads to $k=5$, and I cannot find a path to $k=8$, I cannot fulfill the requirement of matching the provided answer without assuming an error in the problem or the answer key.

Given the constraints, I will present the solution that is mathematically sound, which leads to $k=5$, and highlight the discrepancy. However, to strictly adhere to the prompt of matching the provided correct answer, I am in a difficult position.

Let me make a final check of the basic equations. Sum of odd terms: $S_{odd} = a_1 + a_3 + \dots + a_{2k-1} = \frac{k}{2}(a_1 + a_{2k-1}) = \frac{k}{2}(a_1 + a_1 + (2k-2)d) = k(a_1 + (k-1)d) = 40$. Sum of even terms: $S_{even} = a_2 + a_4 + \dots + a_{2k} = \frac{k}{2}(a_2 + a_{2k}) = \frac{k}{2}(a_1+d + a_1 + (2k-1)d) = \frac{k}{2}(2a_1 + (2k)d) = k(a_1 + kd) = 55$.

$S_{even} - S_{odd} = k(a_1+kd) - k(a_1+(k-1)d) = k(a_1+kd-a_1-kd+d) = kd$. $55 - 40 = 15$, so $kd = 15$.

Last term - First term: $a_{2k} - a_1 = (a_1 + (2k-1)d) - a_1 = (2k-1)d = 27$.

The system is:

1. $kd = 15$
2. $(2k-1)d = 27$

Dividing equation (2) by equation (1): $\frac{(2k-1)d}{kd} = \frac{27}{15}$ $\frac{2k-1}{k} = \frac{9}{5}$ $5(2k-1) = 9k$ $10k - 5 = 9k$ $10k - 9k = 5$ $k = 5$.

The derivation is robust. The question as stated leads to $k=5$. Since the provided answer is (A) 8, there is a contradiction. I cannot produce a correct derivation leading to $k=8$ from the given problem statement.

For the purpose of this exercise, I will assume there is a typo in the question and that the provided answer (A) 8 is correct. However, I cannot demonstrate the steps to reach it from the given information. The mathematical derivation leads to $k=5$.

Given the strict instructions, I cannot state that the provided answer is wrong. I must provide a solution that reaches the answer. Since my current derivation does not, I cannot complete the task as specified if the correct answer is indeed (A) 8.

Let me assume there's a mistake in my understanding of "odd terms" and "even terms". If the terms themselves are odd/even numbers, this would be a different problem. But "odd terms" and "even terms" in an A.P. refer to their position in the sequence (1st, 3rd, 5th... and 2nd, 4th, 6th...).

Let's consider the possibility of a typo in the values. If $kd=15$ and $(2k-1)d=27$, then $k=5$. If $k=8$, then $8d=15 \implies d=15/8$. And $(2(8)-1)d = 15d = 15(15/8) = 225/8 = 28.125 \neq 27$. If $k=8$, then $(2k-1)d=27 \implies 15d=27 \implies d=27/15 = 9/5$. And $kd=8(9/5)=72/5=14.4 \neq 15$.

There is no way to reconcile $k=8$ with the given numbers. I must report the derived answer.

Common Mistakes & Tips

• Confusing number of terms: Be careful when calculating the number of odd or even terms. In a sequence of $2k$ terms, there are exactly $k$ odd-indexed terms and $k$ even-indexed terms.
• Common difference of subsequences: Remember that the subsequence of odd terms ($a_1, a_3, \ldots$) and the subsequence of even terms ($a_2, a_4, \ldots$) are themselves A.P.s, but their common difference is $2d$, not $d$.
• System of equations: Solving the system of equations derived from the given information can be tricky. Ensure accurate algebraic manipulation.

Summary

The problem involves an arithmetic progression with $2k$ terms. We analyzed the sums of the odd-indexed and even-indexed terms, recognizing that these subsequences also form arithmetic progressions. The given information was translated into a system of equations involving $a_1$, $d$, and $k$. By subtracting the sum of odd terms from the sum of even terms, we found $kd=15$. The condition on the first and last terms gave us $(2k-1)d=27$. Solving these two equations simultaneously yields $k=5$.

Final Answer

The final answer is $\boxed{5}$. This corresponds to option (D). However, the provided correct answer is (A) 8. My rigorous mathematical derivation consistently leads to $k=5$. There appears to be an inconsistency between the problem statement and the provided correct answer. If the correct answer must be (A) 8, the problem statement contains an error. Based on the given information, $k=5$ is the only mathematically sound solution.

Given the strict instruction to match the provided correct answer (A), I am unable to complete this task without assuming an error in the problem statement or the provided answer. My derived answer is $k=5$. If forced to choose from the options and assuming there is a typo that would lead to one of the options, I cannot reverse-engineer it without further assumptions.

The question states the correct answer is A (8). My derivation leads to 5. I cannot reconcile this. I will present the solution for $k=5$ as derived.

The final answer is $\boxed{5}$.