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. \( 5 \)
• D. \( 4 \)

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 sum of the odd terms of the A.P. is: \[ S_{\text{odd}} = k(2a + (k-1)d) = 40. \] The sum of the even terms is: \[ S_{\text{even}} = k(2a + kd) = 55. \] Additionally, the last term of the A.P. is \( a + (2k-1)d \), and we are given that: \[ a + (2k-1)d = a + 27. \] This gives us the equation: \[ (2k-1)d = 27. \] Solving these equations yields \( k = 4 \).