Question

If the sum of first $p,q,r$ term of an $a,b,c$ respectively. Prove that $\dfrac{a}{p}\left( {q - r} \right) + \dfrac{b}{q}\left( {r - p} \right) + \dfrac{c}{r}\left( {p - q} \right) = 0$

Answer 1

Hint: We will write the sum of first $p,q,r$ terms using the formula, $\dfrac{n}{2}\left( {2A + \left( {n - 1} \right)D} \right)$ where $A$ is first term of the A.P. and $D$ is the common difference. Then, substitute the values of $\dfrac{a}{p}$, $\dfrac{b}{q}$ and $\dfrac{c}{r}$ in the left-hand-side of the given equation. Simplify it to make it equal to RHS.

Complete step-by-step answer:
We are given that sum of first $p,q,r$ term of an $a,b,c$
We know that the sum of first $n$ terms of a sequence is $\dfrac{n}{2}\left( {2A + \left( {n - 1} \right)D} \right)$, where $A$ is first term of the A.P. and $D$ is the common difference.
Let $A$ be the first term of the A.P. and let $D$ be the common difference of the A.P.
Hence,
$
  a = \dfrac{p}{2}\left( {2A + \left( {p - 1} \right)D} \right) \\
   \Rightarrow \dfrac{a}{p} = \dfrac{1}{2}\left( {2A + \left( {p - 1} \right)D} \right) \\

 $
Similarly, sum of first $q$terms is $b$
$
  b = \dfrac{q}{2}\left( {2A + \left( {q - 1} \right)D} \right) \\
   \Rightarrow \dfrac{b}{q} = \dfrac{1}{2}\left( {2A + \left( {q - 1} \right)D} \right) \\
$
And sum of first $r$ terms is $c$
$
  c = \dfrac{r}{2}\left( {2A + \left( {r - 1} \right)D} \right) \\
   \Rightarrow \dfrac{c}{r} = \dfrac{1}{2}\left( {2A + \left( {r- 1} \right)D} \right) \\
$
We have to prove $\dfrac{a}{p}\left( {q - r} \right) + \dfrac{b}{q}\left( {r - p} \right) + \dfrac{c}{r}\left( {p - q} \right) = 0$
We will substitute the values of $\dfrac{a}{p}$, $\dfrac{b}{q}$ and $\dfrac{c}{r}$ in the above equation.
$\dfrac{1}{2}\left( {2A + \left( {p - 1} \right)D} \right)\left( {q - r} \right)+ \dfrac{1}{2}\left( {2A + \left( {q - 1} \right)D} \right)\left( {r - p} \right)+\dfrac{1}{2}\left( {2A + \left( {r- 1} \right)D} \right)\left( {p - q} \right) =0 $
We will simplify the LHS and will prove it equal to RHS.
$
= \left( {2A + \left( {p - 1} \right)D} \right)\left( {q - r} \right) + \left( {2A + \left( {q - 1} \right)D} \right)\left( {r - p} \right) + \left( {2A + \left( {r - 1} \right)D} \right)\left( {p - q} \right) \\
= \left( {2A + pD - D} \right)\left( {q - r} \right) + \left( {2A + qD - D} \right)\left( {r - p} \right) + \left( {2A + rD - D} \right)\left( {p - q} \right) \\
= 2Aq + pqD - rD - 2rA - rpD + Dr + 2Ar + qrD + rD - 2Ap - pqD + Dp + 2Ap \\
   - 2Aq + rDp - qrD - pD + qD \\
$
On adding and subtracting like terms, we will be left with 0, which is equal to RHS.
Hence, proved.

Note: In an A.P., the sequence is of type $a,a + d,a + 2d,a + 3d,......$. The ${n^{th}}$ term of the sequence is given as ${a_n} = a + \left( {n - 1} \right)d$. And the sum of $n$ terms of sequence is given by $\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ and $\dfrac{n}{2}\left( {a + {a_n}} \right)$