Let the A.P. be $\begin{aligned}
& a, a+2, a+2 d, \ldots, a+(2 k-1) d \
& \text { Now, } a+a+2 d+a+4 d+\ldots+a+(2 k-2) d=40 \
& k a+2 d+4 d+\ldots+(2 k-2) d=40 \
& \Rightarrow k a+\frac{k-1}{2}[2 d+2 k d-2 d]=40
\end{aligned}\Rightarrow k a+k(k-1) d=40...(1)Anda+d+a+3 d+\ldots+a+(2 k-1) d=55\Rightarrow k a+\frac{k}{2}(d+2 k d-d)=55\Rightarrow k a+k^2 d=55....(2)Also,a+(2 k-1) d-a=27\Rightarrow \quad(2 k-1) d=27 \Rightarrow d=\frac{27}{2 k-1}...(3)Fromequation(1)and(2)k^2 d-k d-k^2 d=-15\Rightarrow \quad d=\frac{15}{k}Fromequation(3)and(4)\begin{aligned}
& \frac{27}{2 k-1}=\frac{15}{k} \
& 27 k=30 k-15 \
& \Rightarrow 3 k=15 \
& \Rightarrow k=5
\end{aligned}$
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…
JEE Main 2025 — Mathematics Algebra
Suppose that the number of terms in an A.P. is 2k,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 :
Held on 22 Jan 2025 · Verified 6 Jul 2026.
Options
- A
6
- B
5
- C
8
- D
4
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