A series whose `n^(th)` term is `(n/x)`, then sum of r terms will be

Options

A. `{(r(r+1))/(2x)}+ry`
B. `{(r(r-1))/(2x)}`
C. `{(r(r-1))/(2x)}-xy`
D. `{(r(r+1))/(2x)}-rx`

Answer

To find the sum of the first \( r \) terms of the series whose \( n^{th} \) term is given by \( T_n = \frac{n}{x} + y \), we can follow these steps:

### Step 1: Identify the first term
The first term \( T_1 \) can be calculated by substituting \( n = 1 \) into the formula for \( T_n \):
\[ T_1 = \frac{1}{x} + y \]

### Step 2: Identify the second term
The second term \( T_2 \) can be calculated by substituting \( n = 2 \):
\[ T_2 = \frac{2}{x} + y \]

### Step 3: Identify the third term
The third term \( T_3 \) can be calculated by substituting \( n = 3 \):
\[ T_3 = \frac{3}{x} + y \]

### Step 4: Calculate the common difference
To check if the series is an arithmetic progression (AP), we calculate the common difference \( d \):
\[ d = T_2 - T_1 = \left(\frac{2}{x} + y\right) - \left(\frac{1}{x} + y\right) = \frac{2}{x} - \frac{1}{x} = \frac{1}{x} \]
The common difference is constant, confirming that the series is indeed an AP.

### Step 5: Use the formula for the sum of the first \( r \) terms of an AP
The formula for the sum \( S_r \) of the first \( r \) terms of an arithmetic progression is given by:
\[ S_r = \frac{r}{2} \left(2a + (r - 1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 6: Substitute the values into the formula
Here, \( a = T_1 = \frac{1}{x} + y \) and \( d = \frac{1}{x} \):
\[ S_r = \frac{r}{2} \left(2\left(\frac{1}{x} + y\right) + (r - 1)\left(\frac{1}{x}\right)\right) \]

### Step 7: Simplify the expression
Now, simplify the expression inside the parentheses:
\[ = \frac{r}{2} \left(\frac{2}{x} + 2y + \frac{r - 1}{x}\right) \]
Combine the terms:
\[ = \frac{r}{2} \left(\frac{2 + r - 1}{x} + 2y\right) = \frac{r}{2} \left(\frac{r + 1}{x} + 2y\right) \]

### Final Result
Thus, the sum of the first \( r \) terms is:
\[ S_r = \frac{r}{2} \left(\frac{r + 1}{x} + 2y\right) \]