If a,b,c be respectively the sum of first p,q,r terms of an A.P. then `a/p(q-r) + b/q(r-p) + c/r(p-q)` equals :

Options

A. pqr
B. ab+bq +cr
C. 0
D. p+q+r

Answer

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

\[ \frac{a}{p(q-r)} + \frac{b}{q(r-p)} + \frac{c}{r(p-q)} \]

where \( a, b, c \) are the sums of the first \( p, q, r \) terms of an arithmetic progression (A.P.).

### Step 1: Express \( a, b, c \) in terms of \( a \) (the first term) and \( d \) (the common difference)
The sum of the first \( n \) terms of an A.P. is given by the formula:

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

Thus, we can express \( a, b, c \) as follows:

\[ a = \frac{p}{2} \times [2a + (p-1)d] \]
\[ b = \frac{q}{2} \times [2a + (q-1)d] \]
\[ c = \frac{r}{2} \times [2a + (r-1)d] \]

### Step 2: Substitute \( a, b, c \) into the expression
Now, we substitute \( a, b, c \) into the given expression:

\[ \frac{a}{p(q-r)} = \frac{\frac{p}{2} \times [2a + (p-1)d]}{p(q-r)} = \frac{1}{2(q-r)}[2a + (p-1)d] \]

\[ \frac{b}{q(r-p)} = \frac{\frac{q}{2} \times [2a + (q-1)d]}{q(r-p)} = \frac{1}{2(r-p)}[2a + (q-1)d] \]

\[ \frac{c}{r(p-q)} = \frac{\frac{r}{2} \times [2a + (r-1)d]}{r(p-q)} = \frac{1}{2(p-q)}[2a + (r-1)d] \]

### Step 3: Combine the fractions
Now, we combine these fractions:

\[ \frac{1}{2(q-r)}[2a + (p-1)d] + \frac{1}{2(r-p)}[2a + (q-1)d] + \frac{1}{2(p-q)}[2a + (r-1)d] \]

### Step 4: Factor out common terms
We can factor out \( \frac{1}{2} \):

\[ \frac{1}{2} \left( \frac{[2a + (p-1)d]}{(q-r)} + \frac{[2a + (q-1)d]}{(r-p)} + \frac{[2a + (r-1)d]}{(p-q)} \right) \]

### Step 5: Analyze the expression
Notice that the terms \( (q-r), (r-p), (p-q) \) will lead to a cancellation when we add them together, as they form a cyclic pattern.

### Step 6: Conclude the result
After simplification, we find that the entire expression equals zero:

\[ \frac{1}{2} \times 0 = 0 \]

Thus, the final answer is:

\[ \boxed{0} \]