If `S_1 ,S_2` and `S_3` denote the sum of first `n_1,n_2` and `n_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)` equals :

Options

A. 0
B. 1
C. `S_1S_2S_3`
D. `n_1n_2n_3`

Answer

To solve the problem, we need to evaluate 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 \( S_1, S_2, S_3 \) are the sums of the first \( n_1, n_2, n_3 \) terms of an arithmetic progression (A.P.).

### Step 1: Write the formulas for \( S_1, S_2, S_3 \)

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

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

where \( a \) is the first term and \( d \) is the common difference.

Thus, we can express \( S_1, S_2, S_3 \) as follows:

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

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

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

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

Now we substitute these expressions into the original equation:

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

This simplifies to:

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

### Step 3: Factor out \(\frac{1}{2}\)

Factoring out \(\frac{1}{2}\):

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

### Step 4: Combine the fractions

Now we need to combine these fractions. The common denominator will be \((n_2 - n_3)(n_3 - n_1)(n_1 - n_2)\).

After combining and simplifying, we will notice that the terms will cancel out due to symmetry in the expressions.

### Step 5: Conclusion

After simplification, we find that the entire expression equals \( 0 \).

Thus, the final answer is:

\[ \boxed{0} \]