Question

The sum of first $p,q,r$ terms of an A.P. are $a,b,c$ respectively.
Show 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

We will first write the sum of first $p,q,r$ terms of an A.P, using the formula, ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference. Next, solve the LHS by substituting the required values and prove it equal to RHS.

Complete step-by-step answer:
The sum of first $n$ terms of an A.P. is given by ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference.
Then, the sum of first $p$ terms is
$\Rightarrow {S_p} = \dfrac{p}{2}\left( {2a + \left( {p - 1} \right)d} \right) \\\ \Rightarrow 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, the sum of first $q$ terms of an A.P. is given as
$\Rightarrow {S_q} = \dfrac{q}{2}\left( {2a + \left( {q - 1} \right)d} \right) \\\ \Rightarrow 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 the sum of first $r$ terms of an A.P is given as
$\Rightarrow {S_r} = \dfrac{r}{2}\left( {2a + \left( {r - 1} \right)d} \right) \\\ \Rightarrow 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 LHS of the equation.
$\Rightarrow \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) \\\ \Rightarrow \dfrac{1}{2}\left( {\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)} \right) \\\$
Now, we will open the brackets.
$\Rightarrow \dfrac{1}{2}\left( {\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)} \right) \\\ \Rightarrow \dfrac{1}{2}\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) \\\ \Rightarrow \dfrac{1}{2}\left( {2aq - 2ar + pdq - pdr - dq + dr + 2ar - 2ap + qdr - qdp - dr + pr + 2ap - 2aq + rdp - rdq - dp + dq} \right) \\\ \Rightarrow \dfrac{1}{2}\left( 0 \right) \\\ \Rightarrow 0 \\\$
As LHS=RHS, the given statement is true.

Note: In an A.P. , the sum of first $n$ terms is given as ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference. Also, if the ${n^{th}}$ term is known, then the sum can also be given as ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$, where ${a_n}$ is the last term of the A.P.