JEE Mains · Maths · STD 11 - 8. sequence and series
Suppose that the number of terms in an A.P. is \(2 k, k \in N\). 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 6
- B 5
- C 8
- D 4
Answer & Solution
Correct Answer
(B) 5
Step-by-step Solution
Detailed explanation
Let the A.P. be \(\begin{aligned} & a, a+2, a+2 d, \ldots, a+(2 k-1) d \\ & \text { Now, } a+a+2 d+a+4 d+\ldots+a+(2 k-2) d=40 \\ & k a+2 d+4 d+\ldots+(2 k-2) d=40 \\ & \Rightarrow k a+\frac{k-1}{2}[2 d+2 k d-2 d]=40 \end{aligned}\) \(\Rightarrow k a+k(k-1) d=40\) ...(1) And…
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