JEE Main 2026 Jan 24 Shift 1, Mathematics Q4: Arithmetic, Geometric And Harmonic Progressions

The sequence is a GP with first term $3^6$ and common ratio $3^{-2}$, so the $k$-th term equals $3^{8-2k}$.

$P_n$ multiplies the first $n$ terms; the exponents form the AP $6,4,2,\dots ,8-2n$. Their sum is $n(7-n)$, so

$P_n=3^{n(7-n)}\Rightarrow (P_n{)}^{1/n}=3^{7-n}$.

Hence $\sum _{n=1}^{40}(P_n{)}^{1/n}=\sum _{n=1}^{40}{3}^{7-n}$ is a GP with first term $3^6$, ratio $1/3$, 40 terms:

$\text{Sum}=3^6\cdot \frac{1-3^{-40}}{1-1/3}=\frac{3^7}{2}(1-3^{-40})=\frac{3^{40}-1}{2\cdot 3^{33}}$.

Doubling: $2\sum =\frac{3^{40}-1}{3^{33}}$.

So $\alpha =40$, $\beta =33$, $\gcd (40,33)=1$ ✓. $\therefore \alpha +\beta =73$.