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:
Show 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.
Let \( a \) be the first term and \( d \) be the common difference of the A.P. The sum of the odd terms is \( S_{\text{odd}} = k(2a + (k-1)d) = 40 \). The sum of the even terms is \( S_{\text{even}} = k(2a + kd) = 55 \). The last term of the A.P. is \( a + (2k-1)d \). Given \( a + (2k-1)d = a + 27 \), we have \( (2k-1)d = 27 \). Solving these equations results in \( k = 4 \).
