Question

The sum of first p, q, r terms of an A.P are a, b, c respectively. Show that : \[\dfrac{a}{p}(q - r) + \dfrac{b}{q}(r - p) + \dfrac{c}{r}(p - q) = 0\]

Answer 1

Hint: Use the formula of sum of n terms as \[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\], form all the three equation using the sum of first p, q, r terms of an A.P are a, b, c and equate them to obtain desired result.

Complete step by step answer:
Let the first term of an A.P be A and a common difference be D.
Given that the sum of the first p, q, r terms of an A.P are a, b, c respectively
Now, using the given data we can form all the three equation as
Using the above-mentioned concept in the hint we can move for the formation of three equations for the given term using the mentioned first term and common difference.
\[
  a = \dfrac{p}{2}(2A + (p - 1)D) \\
  b = \dfrac{q}{2}(2A + (q - 1)D) \\
  c = \dfrac{r}{2}(2A + (r - 1)D) \\
 \]
Now, simplify the above equations all the three equations, and mold it in such a way that all the three equations are somewhat similar and then subtract them to cancel out known terms and to form the equation only with known values.
\[
  \dfrac{a}{p} = A + \dfrac{{(p - 1)D}}{2} \\
  \dfrac{b}{q} = A + \dfrac{{(q - 1)D}}{2} \\
  \dfrac{c}{r} = A + \dfrac{{(r - 1)D}}{2} \\
 \]
Now, calculating the value of \[\dfrac{a}{p}(q - r)\]
\[
  \dfrac{a}{p}(q - r) = [A + \dfrac{{(p - 1)D}}{2}](q - r) \\
   = A(q - r) + \dfrac{{(p - 1)D}}{2}(q - r) \\
 \]
Now, similarly calculate the value of \[\dfrac{b}{q}(r - p)\]
\[\dfrac{b}{q}(r - p) = A(r - p) + \dfrac{{(q - 1)}}{2}D(r - p)\]
And also similarly for , \[\dfrac{c}{r}(p - q) = A(p - q) + \dfrac{{(q - 1)}}{2}D(p - q)\]
Now put all the above obtained values in , \[\dfrac{a}{p}(q - r) + \dfrac{b}{q}(r - p) + \dfrac{c}{r}(p - q)\]
\[
   \Rightarrow A(p - q) + \dfrac{{(q - 1)}}{2}D(p - q) + A(q - r) + \dfrac{{(p - 1)D}}{2}(q - r) + A(r - p) + \dfrac{{(q - 1)}}{2}D(r - p) \\
   \Rightarrow A[(p - q) + (q - r) + (r - p)] + \dfrac{D}{2}[(p - q)(q - 1) + (q - r)(p - 1) + (r - p)(q - 1)] \\
 \]
Keeping on simplifying it further,
\[
   = A[p - q + q - r + r - p] + \dfrac{D}{2}[(pq - pr - q + qr - pq - r + p + rp - rq - p + q] \\
   = A[0] + D[0] \\
   = 0 \\
 \]
Hence, \[\dfrac{a}{p}(q - r) + \dfrac{b}{q}(r - p) + \dfrac{c}{r}(p - q) = 0\].

Note: In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
The formula for calculation of sum of n terms of an A.P as \[{s_n} = \dfrac{n}{2}(2a + (n - 1)d)\]. Use the given values carefully and substitute them in such a way and cancel the terms which will lead us to prove L.H.S=R.H.S.