JEE Main 2025 Apr 4 Shift 2, Mathematics Q21: Arithmetic, Geometric And Harmonic Progressions

Take $\alpha =\omega$, a primitive cube root of unity, so ${\omega }^3=1$ and $1/{\omega }^k={\omega }^{-k}$. Then

${\left({\omega }^k+{\omega }^{-k}\right)}^2={\omega }^{2k}+2+{\omega }^{-2k}={\omega }^{2k}+{\omega }^k+2$,

using ${\omega }^{-2k}={\omega }^k$ (since ${\omega }^{3k}=1$).

So $\sum _{k=1}^n({\omega }^{2k}+{\omega }^k+2)=2n+\sum _{k=1}^n{\omega }^k+\sum _{k=1}^n{\omega }^{2k}=20$.

The sums of ${\omega }^k$ in groups of 3 cancel because $1+\omega +{\omega }^2=0$. Write $n=3m+r$ with $r\in \{0,1,2\}$ and look at the remainder.

$r=0$: both sums are $0$, giving $2n=20\Rightarrow n=10$, but $10$ is not a multiple of $3$. Reject.

$r=1$: residual is $\omega +{\omega }^2=-1$, giving $2n-1=20\Rightarrow n=21/2$. Reject.

$r=2$: residual is $(\omega +{\omega }^2)+({\omega }^2+{\omega }^4)=-1+({\omega }^2+\omega )=-2$. So $2n-2=20\Rightarrow n=11$, and $11=3(3)+2$. Accept.

$\therefore n=11$.