JEE Main 2025 — Mathematics Algebra

$$\begin{aligned} & \alpha=\omega \\ & \therefore\left(\omega^k+\frac{1}{\omega^k}\right)^2=\omega^{2 k}+\frac{1}{\omega^{2 k}}+2 \\ & =\omega^{2 k}+\omega^k+2 \quad \because \omega^{3 k}=1\end{aligned}$$​α=ω∴(ωk+ωk1​)2=ω2k+ω2k1​+2=ω2k+ωk+2∵ω3k=1​

$\begin{aligned}
& \therefore \sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\omega^{2 \mathrm{k}}+\omega^{\mathrm{k}}+2\right)=20 \
& \Rightarrow\left(\omega^2+\omega^4+\omega^6+\ldots+\omega^{2 \mathrm{n}}\right)+\left(\omega+\omega^2+\omega^3+\ldots+\right. \
& \left.\omega^{\mathrm{n}}\right)+2 \mathrm{n}=20
\end{aligned}$

Now if $n=3 m, \quad m \in I$n=3m,m∈I

Then $0+0+2 \mathrm{n}=20 \Rightarrow \mathrm{n}=10$0+0+2n=20⇒n=10 (not satisfy)

if $n=3 m+1$n=3m+1, then

$\begin{aligned}
& \omega^2+\omega+2 \mathrm{n}=20 \
& -1+2 \mathrm{n}=20 \Rightarrow \mathrm{n}=\frac{21}{2}(\text { not possible })
\end{aligned}$

if $n=3 m+2$n=3m+2,

$\begin{aligned}
& \left(\omega^8+\omega^{10}\right)+\left(\omega^4+\omega^5\right)+2 \mathrm{n}=20 \
& \Rightarrow\left(\omega^2+\omega\right)+\left(\omega+\omega^2\right)+2 \mathrm{n}=20 \
& 2 \mathrm{n}=22 \
& \mathrm{n}=11 \text { satisfy } \mathrm{n}=3 \mathrm{~m}+2 \
& \therefore \mathrm{n}=11
\end{aligned}$