(i) If the ratio of sum of first n terms of two A.P. s in `(7n+1):(4n:27)`, then ratio of their `m^(th)` terms is `ul("P")`.
(ii) Sum of n odd natural numbers is `ul("Q")`.
(iii) If sum of first n terms of three A.P.s are `S_(1), S_(2), S_(3)`. The first term of each is 1 and common difference are 1, 2 and 3 respectively, then `(S_(1)+S_(3))/(S_(2))=ul("R")`.

Options

A. `{:("P","Q","R"),((14m-6)/(8m+23),n^(2),2):}`
B. `{:("P","Q","R"),((13m+6)/(7m+9),n^(2),5):}`
C. `{:("P","Q","R"),((14m+6)/(8m+23),2n+1,1):}`
D. `{:("P","Q","R"),((7m+1)/(4m+27),2n-1,3):}`

Answer

To solve the given problem, we will break it down into three parts as stated in the question.

### Part (i): Ratio of the m-th terms of two A.P.s

1. **Given Information**: The ratio of the sum of the first n terms of two A.P.s is given as:
\[ S_1 : S_2 = (7n + 1) : (4n + 27) \]
The formula for the sum of the first n terms of an A.P. is:
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Expressing the Sums**:
Let the first A.P. have first term \( a_1 \) and common difference \( d_1 \), and the second A.P. have first term \( a_2 \) and common difference \( d_2 \).
\[ S_1 = \frac{n}{2} [2a_1 + (n-1)d_1] \]
\[ S_2 = \frac{n}{2} [2a_2 + (n-1)d_2] \]

3. **Setting Up the Ratio**:
\[ \frac{S_1}{S_2} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n + 1}{4n + 27} \]

4. **Finding the m-th Terms**:
The m-th term of an A.P. is given by:
\[ T_m = a + (m-1)d \]
Therefore, for both A.P.s:
\[ T_{1m} = a_1 + (m-1)d_1 \]
\[ T_{2m} = a_2 + (m-1)d_2 \]

5. **Finding the Ratio of m-th Terms**:
The ratio of the m-th terms is:
\[ \frac{T_{1m}}{T_{2m}} = \frac{a_1 + (m-1)d_1}{a_2 + (m-1)d_2} = P \]

### Part (ii): Sum of n odd natural numbers

1. **Formula for Sum of Odd Natural Numbers**:
The sum of the first n odd natural numbers is given by:
\[ Q = n^2 \]

### Part (iii): Ratio of sums of three A.P.s

1. **Given Information**: The first terms of all three A.P.s are 1, and their common differences are 1, 2, and 3 respectively.

2. **Finding Sums**:
- For the first A.P.:
\[ S_1 = \frac{n}{2} [2 \cdot 1 + (n-1) \cdot 1] = \frac{n}{2} [2 + n - 1] = \frac{n(n + 1)}{2} \]
- For the second A.P.:
\[ S_2 = \frac{n}{2} [2 \cdot 1 + (n-1) \cdot 2] = \frac{n}{2} [2 + 2n - 2] = n^2 \]
- For the third A.P.:
\[ S_3 = \frac{n}{2} [2 \cdot 1 + (n-1) \cdot 3] = \frac{n}{2} [2 + 3n - 3] = \frac{n(3n - 1)}{2} \]

3. **Finding the Ratio**:
\[ \frac{S_1 + S_3}{S_2} = \frac{\frac{n(n + 1)}{2} + \frac{n(3n - 1)}{2}}{n^2} \]
Simplifying:
\[ = \frac{n(n + 1 + 3n - 1)}{2n^2} = \frac{n(4n)}{2n^2} = \frac{4}{2} = 2 \]

### Final Answers:
- \( P \): Ratio of m-th terms of two A.P.s (to be determined from part (i)).
- \( Q = n^2 \)
- \( R = 2 \)