Sequence and Series – Geometric Progression – JEE Main 24 Jan 2026 Shift 1

Solution:

The given sequence is a Geometric Progression (G.P.): $729, 81, 9, 1, \dots$
First term $a = 729 = 3^6$.
Common ratio $r = \frac{81}{729} = \frac{1}{9} = 3^{-2}$.

The $k$-th term of the sequence is $T_k = a r^{k-1} = 3^6 \cdot (3^{-2})^{k-1} = 3^{6 – 2k + 2} = 3^{8-2k}$.

The product of the first $n$ terms is:
$$P_n = \prod_{k=1}^{n} T_k = \prod_{k=1}^{n} 3^{8-2k} = 3^{\sum_{k=1}^{n} (8-2k)}$$

Calculating the exponent:
$$\sum_{k=1}^{n} (8-2k) = 8n – 2 \sum_{k=1}^{n} k = 8n – 2\frac{n(n+1)}{2} = 8n – n^2 – n = 7n – n^2$$
So, $P_n = 3^{7n – n^2}$.

We need to evaluate $(P_n)^{\frac{1}{n}}$:
$$(P_n)^{\frac{1}{n}} = (3^{n(7-n)})^{\frac{1}{n}} = 3^{7-n}$$

Now, substitute this into the given summation:
$$S = 2 \sum_{n=1}^{40} 3^{7-n}$$
$$S = 2 (3^6 + 3^5 + 3^4 + \dots + 3^{7-40})$$
The series inside the bracket is a G.P. with first term $A = 3^6$, common ratio $R = \frac{1}{3}$, and number of terms $N = 40$.

Sum of G.P. $= A \frac{1-R^N}{1-R}$:
$$\sum_{n=1}^{40} 3^{7-n} = 3^6 \frac{1 – (\frac{1}{3})^{40}}{1 – \frac{1}{3}} = 3^6 \frac{1 – \frac{1}{3^{40}}}{\frac{2}{3}}$$
$$= \frac{3^7}{2} \left( \frac{3^{40}-1}{3^{40}} \right) = \frac{1}{2} \cdot \frac{3^{40}-1}{3^{33}}$$

Therefore, the total expression is:
$$2 \times \frac{1}{2} \frac{3^{40}-1}{3^{33}} = \frac{3^{40}-1}{3^{33}}$$

Comparing with $\frac{3^{\alpha}-1}{3^{\beta}}$:
$\alpha = 40$ and $\beta = 33$.
gcd(40, 33) = 1 (Verified).

We need $\alpha + \beta$:
$$\alpha + \beta = 40 + 33 = 73$$

Ans. (1)