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

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.