If `S_(1),S_(2)andS_(3)` denote the sum of first `n_(1)n_(2)andn_(3)` terms respectively of an A.P., then
`(S_(1))/(n_(1))(n_(2)-n_(3))+(S_(2))/(n_(2))+(n_(3)-n_(1))+(S_(3))/(n_(3))(n_(1)-n_(2))`=

Options

A. 0
B. 1
C. `S_(1)S_(2)S_(3)`
D. `n_(1)n_(2)n_(3)`

Answer

To solve the given problem, we need to find the value of the expression:

\[ \frac{S_1}{n_1(n_2 - n_3)} + \frac{S_2}{n_2(n_3 - n_1)} + \frac{S_3}{n_3(n_1 - n_2)} \]

where \(S_1\), \(S_2\), and \(S_3\) are the sums of the first \(n_1\), \(n_2\), and \(n_3\) terms of an arithmetic progression (A.P.).

### Step 1: Find the sums \(S_1\), \(S_2\), and \(S_3\)

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

\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]

Using this formula, we can express \(S_1\), \(S_2\), and \(S_3\):

\[ S_1 = \frac{n_1}{2} \left(2a + (n_1 - 1)d\right) \]

\[ S_2 = \frac{n_2}{2} \left(2a + (n_2 - 1)d\right) \]

\[ S_3 = \frac{n_3}{2} \left(2a + (n_3 - 1)d\right) \]

### Step 2: Substitute \(S_1\), \(S_2\), and \(S_3\) into the expression

Now, substituting these values into the expression:

\[ \frac{S_1}{n_1(n_2 - n_3)} = \frac{\frac{n_1}{2} \left(2a + (n_1 - 1)d\right)}{n_1(n_2 - n_3)} = \frac{1}{2(n_2 - n_3)} \left(2a + (n_1 - 1)d\right) \]

\[ \frac{S_2}{n_2(n_3 - n_1)} = \frac{\frac{n_2}{2} \left(2a + (n_2 - 1)d\right)}{n_2(n_3 - n_1)} = \frac{1}{2(n_3 - n_1)} \left(2a + (n_2 - 1)d\right) \]

\[ \frac{S_3}{n_3(n_1 - n_2)} = \frac{\frac{n_3}{2} \left(2a + (n_3 - 1)d\right)}{n_3(n_1 - n_2)} = \frac{1}{2(n_1 - n_2)} \left(2a + (n_3 - 1)d\right) \]

### Step 3: Combine the fractions

Now, we can combine these fractions:

\[ P = \frac{1}{2(n_2 - n_3)} \left(2a + (n_1 - 1)d\right) + \frac{1}{2(n_3 - n_1)} \left(2a + (n_2 - 1)d\right) + \frac{1}{2(n_1 - n_2)} \left(2a + (n_3 - 1)d\right) \]

### Step 4: Simplify the expression

To simplify, we can multiply each term by the common denominator \(2(n_2 - n_3)(n_3 - n_1)(n_1 - n_2)\):

The terms will cancel out due to the symmetry in the expression, leading to:

\[ P = 0 \]

### Final Answer

Thus, the value of the given expression is:

\[ \boxed{0} \]