Question

Match the LIST-I with LIST-II

Choose the correct answer from the options given below:

• A. A-IV, B-II, C-III, D-I
• B. A-IV, B-II, C-I, D-III
• C. A-II, B-IV, C-III, D-I
• D. A-II, B-IV, C-I, D-III

Hint:

Always simplify exponential bases first: $4^3 = 64$ and $5^2 = 25$. Comparing $64$ and $25$ immediately reveals whether $r > 1$ or $r < 1$!

Answer 1

The Correct Option is B

Step 1 : Concept:
This question tests limits of geometric sequences and convergence/divergence of geometric series.

Step 2 : Key Formulas and Approach:
1. $4^{3n} = (4^3)^n = 64^n$ and $5^{2n} = (5^2)^n = 25^n$. 2. For $r > 1$: $\lim_{n \to \infty} r^n = \infty$, and $\sum_{n=0}^\infty r^n$ diverges. 3. For $0 \le r < 1$: $\lim_{n \to \infty} r^n = 0$, and $\sum_{n=0}^\infty r^n$ converges.

Step 3 : Step-by-step Explanation:

• Item A:
$x_n = \frac{4^{3n}}{5^{2n}} = \frac{64^n}{25^n} = \left(\frac{64}{25}\right)^n$. Since $r = \frac{64}{25} > 1$, $\lim_{n \to \infty} x_n = \infty$. Matches with IV.

• Item B:
$x_n = \frac{5^{2n}}{4^{3n}} = \frac{25^n}{64^n} = \left(\frac{25}{64}\right)^n$. Since $r = \frac{25}{64} < 1$, $\lim_{n \to \infty} x_n = 0$. Matches with II.

• Item C:
$\sum_{n=0}^{\infty} \frac{4^{3n}}{5^{2n}} = \sum_{n=0}^{\infty} \left(\frac{64}{25}\right)^n$. Since common ratio $r = \frac{64}{25} > 1$, the geometric series is Divergent. Matches with I.

• Item D:
$\sum_{n=0}^{\infty} \frac{5^{2n}}{4^{3n}} = \sum_{n=0}^{\infty} \left(\frac{25}{64}\right)^n$. Since common ratio $r = \frac{25}{64} < 1$, the geometric series is Convergent. Matches with III.

Step 4 : Final Answer:
The correct matching is A-IV, B-II, C-I, D-III, which corresponds to option (B).