Question Detail
Question
Suppose that the number of terms in an A.P. is 2k. If the sum of all odd terms is 40, the sum of all even terms is 55, and the last term exceeds the first term by 27. Find k.
Correct Answer
Option C is the correct answer.
Detailed Explanation
Let first term = a, common difference = d. Number of terms = 2k. Last term = a + (2k-1)d. Given last - first = 27 ⇒ (2k-1)d = 27. Sum of odd terms: terms at positions 1,3,5,...,2k-1. There are k odd terms. These form an AP with first term a, common difference 2d. Sum = (k/2)[2a + (k-1)2d] = k[a + (k-1)d] = 40. Sum of even terms: terms at positions 2,4,...,2k. There are k terms, first term a+d, common difference 2d. Sum = (k/2)[2(a+d) + (k-1)2d] = k[(a+d) + (k-1)d] = k[a + kd] = 55. Subtract: k[a+kd] - k[a+(k-1)d] = 55-40 ⇒ kd = 15 ⇒ d = 15/k. Then (2k-1)(15/k) = 27 ⇒ 15(2k-1) = 27k ⇒ 30k - 15 = 27k ⇒ 3k = 15 ⇒ k = 5.
Hint
Set up equations for the sums of odd and even terms separately.