Sequences and Series – Arithmetico-Geometric Progression – JEE Main 24 January 2025 Shift 2

Solution:

Let the sum of the series be $S$. Given $S=7$.
The series is an Arithmetico-Geometric Progression (AGP).
$$ S = 5 + \frac{5+\alpha}{7} + \frac{5+2\alpha}{7^2} + \frac{5+3\alpha}{7^3} + \dots $$
Multiply the equation by the common ratio $\frac{1}{7}$:
$$ \frac{1}{7}S = \frac{5}{7} + \frac{5+\alpha}{7^2} + \frac{5+2\alpha}{7^3} + \dots $$

Subtracting the second equation from the first:
$$ S – \frac{1}{7}S = 5 + \left( \frac{5+\alpha}{7} – \frac{5}{7} \right) + \left( \frac{5+2\alpha}{7^2} – \frac{5+\alpha}{7^2} \right) + \dots $$
$$ \frac{6}{7}S = 5 + \frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots $$

The terms starting from the second term form an infinite Geometric Progression with first term $a = \frac{\alpha}{7}$ and ratio $r = \frac{1}{7}$.
Sum of GP $= \frac{a}{1-r} = \frac{\alpha/7}{1 – 1/7} = \frac{\alpha/7}{6/7} = \frac{\alpha}{6}$.

So, the equation becomes:
$$ \frac{6}{7}S = 5 + \frac{\alpha}{6} $$
Given $S = 7$:
$$ \frac{6}{7}(7) = 5 + \frac{\alpha}{6} $$
$$ 6 = 5 + \frac{\alpha}{6} $$
$$ 1 = \frac{\alpha}{6} \Rightarrow \alpha = 6 $$

Ans. (3)