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JEE MAIN Mathematics QUESTION #974
Question 1
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 exceeds the first term by 27, then \(k\) is equal to:
  • 6
  • 5✔️
  • 8
  • 4
Correct Answer Explanation
Let first term \(=a\), common difference \(=d\), number of terms \(=2k\). Sum of odd-positioned terms \(=k\left(a+\frac{(2k-1)d}{2}\cdot\text{...}\right)=40\), even terms \(=55\). Difference: \(kd=15\). Last \(-\) first \(=(2k-1)d=27\). From \(kd=15\) and \((2k-1)d=27\): dividing gives \(2k-1=\frac{27k}{15} \Rightarrow k=5\).