Suppose that the number of terms in an A.P. is 2 k, k ∈ 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:

Solution

Maths sir · 4 steps

Pair each even term with the odd term just before it: every pair differs by exactly d.

1. Pair the terms. With 2k terms there are k odd-position terms and k even-position terms. Pair a₂ with a₁, a₄ with a₃, and so on, giving k pairs. In each pair the even term is larger by d.
2. Two equations. So the even sum minus the odd sum is kd = 55 - 40 = 15. The last term is the (2k)th, so last minus first is (2k-1)d = 27.

   (2k-1)d/kd = 27/15 = 9/5 ⇒ 10k - 5 = 9k

3. Check. This gives k = 5 and d = 3. Then the odd sum is 5a + 60 = 40, so a = -4, and the even sum is 5a + 75 = 55, which matches.
4. So. k = 5, which is option (d).

The idea to keep. When an A.P. has an even number of terms, the even-place sum minus the odd-place sum is (number of pairs) × d. Dividing two equations that both contain d removes d completely.

Answer from the NTA final answer key. Solution by PrepWiser.