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 \( a \), \( b \), and \( c \) are the sums of the first \( p \), \( q \), and \( r \) terms of an arithmetic progression (A.P.) respectively.
### Step 1: Write the formulas for the sums of the first \( p \), \( q \), and \( r \) terms of an A.P.
The sum of the first \( n \) terms of an A.P. can be expressed as:
\[
S_n = \frac{n}{2} \left(2a + (n-1)d\right)
\]
where \( a \) is the first term and \( d \) is the common difference.
Thus, we have:
1. For \( p \) terms:
\[
S_p = \frac{p}{2} \left(2A + (p-1)d\right) = a
\]
2. For \( q \) terms:
\[
S_q = \frac{q}{2} \left(2A + (q-1)d\right) = b
\]
3. For \( r \) terms:
\[
S_r = \frac{r}{2} \left(2A + (r-1)d\right) = c
\]
### Step 2: Rearranging the equations
From these equations, we can express \( 2A + (n-1)d \) for \( n = p, q, r \):
1. From \( S_p \):
\[
2A + (p-1)d = \frac{2a}{p}
\]
2. From \( S_q \):
\[
2A + (q-1)d = \frac{2b}{q}
\]
3. From \( S_r \):
\[
2A + (r-1)d = \frac{2c}{r}
\]
### Step 3: Subtracting the equations
Now, we will subtract these equations to find expressions for \( q - r \), \( r - p \), and \( p - q \):
1. Subtract the third equation from the second:
\[
(2A + (q-1)d) - (2A + (r-1)d) = \frac{2b}{q} - \frac{2c}{r}
\]
This simplifies to:
\[
(q - r)d = \frac{2b}{q} - \frac{2c}{r}
\]
2. Subtract the first equation from the third:
\[
(2A + (r-1)d) - (2A + (p-1)d) = \frac{2c}{r} - \frac{2a}{p}
\]
This simplifies to:
\[
(r - p)d = \frac{2c}{r} - \frac{2a}{p}
\]
3. Subtract the second equation from the first:
\[
(2A + (p-1)d) - (2A + (q-1)d) = \frac{2a}{p} - \frac{2b}{q}
\]
This simplifies to:
\[
(p - q)d = \frac{2a}{p} - \frac{2b}{q}
\]
### Step 4: Substitute back into the original expression
Now we substitute these results back into the original expression:
\[
\frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q)
\]
Using the relationships we derived, we can express each term in terms of \( d \) and the sums \( a, b, c \).
### Step 5: Simplifying the expression
After substituting and simplifying, we find that all terms cancel out, leading to:
\[
\frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q) = 0
\]
### Final Answer
Thus, the value of the expression is:
\[
\boxed{0}
\]
