In an A.P. if `S_(1) = T_(1) + T_(2) + T_(3) +…+ T_(n)` (n odd) `S_(2) = T_(1) + T_(3) + T_(5) + …+ T_(n)`, then `S_(1)//S_(2)` =

Options

A. `(2n)/(n + 1)`
B. `(n)/(n + 1)`
C. `(n + 1)/(2n)`
D. `(n + 1)/(n)`

Answer

To solve the problem, we need to find the ratio \( \frac{S_1}{S_2} \) where:

- \( S_1 = T_1 + T_2 + T_3 + \ldots + T_n \)
- \( S_2 = T_1 + T_3 + T_5 + \ldots + T_n \)

Here, \( n \) is an odd number.

### Step 1: Calculate \( S_1 \)

The sum \( S_1 \) is the sum of the first \( n \) terms of an arithmetic progression (A.P.). The formula for the sum of the first \( n \) terms of an A.P. is given by:

\[ S_n = \frac{n}{2} \times (T_1 + T_n) \]

Thus, we have:

\[ S_1 = \frac{n}{2} \times (T_1 + T_n) \]

### Step 2: Calculate \( S_2 \)

Next, we calculate \( S_2 \), which is the sum of the first, third, fifth, and so on, terms of the A.P. Since \( n \) is odd, the number of terms in \( S_2 \) will be:

\[ \text{Number of terms in } S_2 = \frac{n + 1}{2} \]

The last term in \( S_2 \) is \( T_n \) (since \( n \) is odd). The first term is \( T_1 \). Therefore, we can write:

\[ S_2 = \frac{(n + 1)/2}{2} \times (T_1 + T_n) = \frac{n + 1}{4} \times (T_1 + T_n) \]

### Step 3: Find the ratio \( \frac{S_1}{S_2} \)

Now we can find the ratio \( \frac{S_1}{S_2} \):

\[ \frac{S_1}{S_2} = \frac{\frac{n}{2} \times (T_1 + T_n)}{\frac{n + 1}{4} \times (T_1 + T_n)} \]

Since \( (T_1 + T_n) \) is common in both the numerator and the denominator, we can cancel it out (assuming \( T_1 + T_n \neq 0 \)):

\[ \frac{S_1}{S_2} = \frac{n/2}{(n + 1)/4} = \frac{n}{2} \times \frac{4}{n + 1} = \frac{2n}{n + 1} \]

### Final Result

Thus, the final result is:

\[ \frac{S_1}{S_2} = \frac{2n}{n + 1} \]