Quantitative Aptitude

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2020 Q.61)** **1. Governing Concept & Definition**: - **Topic**: Engineering Mathematics & Arithmetico-Geometric Progression (AGP) - The given infinite series is: $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \frac{5}{3^4} + \dots = \sum_{n=1}^\infty \frac{n}{3^{n-1}}$ - Sum of infinite AGP with $a = 1, d = 1, r = 1/3$: $S = \frac{a}{1 - r} + \frac{d r}{(1 - r)^2}$ **2. Detailed Step-by-Step Calculation**: - **Method of Multiplying by Ratio $r = 1/3$**: $S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots$ $\frac{1}{3}S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \dots$ - **Subtracting the two series**: $S - \frac{1}{3}S = 1 + \left(\frac{2}{3} - \frac{1}{3}\right) + \left(\frac{3}{9} - \frac{2}{9}\right) + \left(\frac{4}{27} - \frac{3}{27}\right) + \dots$ $\frac{2}{3}S = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$ - **Summing the infinite geometric series**: $\frac{2}{3}S = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2}$ $S = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4} = 2.25$ **3. Final Answer**: - **Calculated Value**: **2.25** - **Official GATE Accepted Range**: **2.20 to 2.30**