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. contains \( 2k \) terms. The final term of the A.P. is \( a + (2k-1)d \). Given that this final term exceeds the first term by 27, we establish the equation:

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

This equation simplifies to:

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

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

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

This can be expressed as:

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

Similarly, the sum of the even-indexed terms (\( a+d, a+3d, a+5d, \ldots \)) also has \( k \) terms. The sum of these even terms, \( S_{\text{even}} \), is given as 55:

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

This simplifies to:

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

We now have the following system of equations:

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

From the first equation, we isolate \( d \):

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

Substitute \( d \) into the second equation:

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

Substitute \( d \) into the third equation:

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

Subtract the second equation from the third equation:

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

This simplifies to:

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

To solve for \( k \), we rearrange:

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

Simplifying further:

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

This equation resolves to:

\( 30k - 15 = 27k \)

Rearranging yields:

\( 3k = 15 \)

Therefore:

\( k = 5 \)

The value of \( k \) that satisfies all given conditions is 5.