Step-by-step Solution:

Let's solve each problem step by step:

Problem A: Finding the Common Ratio in a Geometric Progression (GP)
Given: Third term = 24, Sixth term = 192.
In a GP, the nth term is given by: \[ a_n = a r^{(n-1)}
\] Third term: \[ a r^2 = 24
\] Sixth term: \[ a r^5 = 192
\] Dividing the second equation by the first: \[ \frac{a r^5}{a r^2} = \frac{192}{24}
\] \[ r^3 = 8
\] \[ r = 2
\] So, the common ratio is 2 → A - IV.
Problem B: Sum of an Arithmetic Progression (AP)
Given: \( S_{2n} = 3n \), we need to find \( \frac{S_{3n}}{S_n} \).
The sum of an AP is given by: \[ S_n = \frac{n}{2} (2a + (n-1)d)
\] Since \( S_{2n} = 3n \), substituting in the formula: \[ \frac{2n}{2} (2a + (2n-1)d) = 3n
\] \[ n(2a + (2n-1)d) = 3n
\] \[ 2a + (2n-1)d = 3
\] Similarly, for \( S_{3n} \): \[ S_{3n} = \frac{3n}{2} (2a + (3n-1)d)
\] \[ S_{3n} = \frac{3n}{2} \times 6
\] \[ S_{3n} = 9n
\] Finding \( \frac{S_{3n}}{S_n} \): \[ \frac{S_{3n}}{S_n} = \frac{9n}{3n} = 3
\] So, the answer is 2 → B - II.
Problem C: Finding the First Term in a GP
Given: Sum of the first 3 terms = \( \frac{13}{12} \), and the product = -1.
Let the terms be \( a, ar, ar^2 \). \[ a + ar + ar^2 = \frac{13}{12}
\] \[ a \cdot ar \cdot ar^2 = -1
\] \[ a^3 r^3 = -1
\] \[ (ar)^3 = -1
\] - Solving this system, we find that a = -1.
So, the answer is C - III.
Problem D: Finding the Least Value of \( n \) for Which the Sum Exceeds 1000
The sum of an arithmetic series is: \[ S_n = \frac{n}{2} (2a + (n-1)d)
\] - Given: First term \( a = 3 \), common difference \( d = 3 \), sum \( S_n > 1000 \). \[ S_n = \frac{n}{2} (6 + (n-1) \cdot 3)
\] \[ S_n = \frac{n}{2} (3n + 3)
\] \[ S_n = \frac{3n(n+1)}{2} > 1000
\] - Solving for \( n \): \[ 3n(n+1) > 2000
\] - Trying \( n = 25 \): \[ 3(25 \times 26) = 3 \times 650 = 1950
\] - Trying \( n = 26 \): \[ 3(26 \times 27) = 3 \times 702 = 2106
\] - So, the least value is n = 26 → D - I.
Final Matching:
A - IV
B - II
C - III
D - I

Correct Answer:
✔ (c) A – IV; B – II; C – III; D – I