Question

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:

• A. \( 8 \)
• B. \( 6 \)
• C. \( 4 \)
• D. \( 5 \)

Hint:

In problems involving arithmetic progressions (APs): - Use the sum formula for arithmetic progressions: \( S_n = \frac{n}{2} \left( 2a + (n-1) d \right) \). - For terms involving odd and even sums, break the AP into odd and even indexed terms and use these formulas separately to simplify the process.

Answer 1

The Correct Option is D

Let the first term of the A.P. be \( a \) and the common difference be \( d \). The A.P. has \( 2k \) terms. The last term of the A.P. can be expressed as \( a + (2k-1)d \). According to the problem, this last term exceeds the first term by 27, so we have:

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

This simplifies to:

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

The sum of the odd terms \( a, a+2d, a+4d, \ldots \) consists of \( k \) terms. Using the sum formula for an A.P., the sum \( S_{\text{odd}} \) of these terms is given as 40:

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

This can be rewritten as:

\( k(a + (k-1)d) = 40 \)

Similarly, the sum of the even terms \( a+d, a+3d, a+5d, \ldots \) also consists of \( k \) terms, and the sum \( S_{\text{even}} \) is given as 55:

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

This simplifies to:

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

Now, we have the following system of equations:

• \( (2k-1)d = 27 \)
• \( k(a + (k-1)d) = 40 \)
• \( k(a + kd) = 55 \)

Simplifying the first equation, we have:

\( d = \frac{27}{2k-1} \)

Substitute \( d \) in the second equation:

\( k(a + (k-1)\frac{27}{2k-1}) = 40 \)

From the third equation, substituting \( d \):

\( k(a + k\frac{27}{2k-1}) = 55 \)

Subtract the second equation from the third:

\( k(k\frac{27}{2k-1} - (k-1)\frac{27}{2k-1}) = 15 \)

This reduces to:

\( k(\frac{27}{2k-1}) = 15 \)

Solving for \( k \), we substitute:

\( k = \frac{15(2k-1)}{27} \)

Simplifying:

\( 15(2k-1) = 27k \)

This simplifies to:

\( 30k - 15 = 27k \)

Therefore:

\( 3k = 15 \)

Thus:

\( k = 5 \)

Thus, the correct value of \( k \) that satisfies all conditions is 5.