Question

If the sum of first p terms, first q terms and first r terms of an A.P be x, y and z respectively, then $\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q)$ is

Answer 1

Hint: Use the sum of n terms formula in A.P which is given by ${{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]$ ; where $a$ is the first term of A.P and d is the common difference. Try to find the value of $\dfrac{x}{p},\dfrac{y}{q},\dfrac{z}{r}$ using the sum of n terms formula. Substitute the values so obtained in $\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q)$ .

Complete step by step answer:
We are given in the question that sum of first p terms ${{S}_{p}}$ , first q terms ${{S}_{q}}$ and first r terms ${{S}_{r}}$ be x, y and z respectively, so
$\left. \begin{align}
  & {{S}_{p}}=x \\
 & {{S}_{q}}=y \\
 & {{S}_{r}}=z \\
\end{align} \right\}......(1)$
We know that sum of n terms in A.P is given by ${{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]$ ; where $a$ is the first term of A.P and d is the common difference. Using this formula of sum of n term and equation (1) can be written as
${{S}_{p}}=\dfrac{p}{2}[2a+(p-1)d]=x......(2)$
${{S}_{q}}=\dfrac{q}{2}[2a+(q-1)d]=y......(3)$
${{S}_{r}}=\dfrac{r}{2}[2a+(r-1)d]=z......(4)$
From equation (2) we will divide p throughout the equation and divide 2 inside the bracket in left hand side of equation we get,
 $\dfrac{x}{p}=a+(p-1)\dfrac{d}{2}......(5)$
Similarly from equation (3) we have,
$\dfrac{y}{q}=a+(q-1)\dfrac{d}{2}......(6)$
Similarly from equation (4) we have,
$\dfrac{z}{r}=a+(r-1)\dfrac{d}{2}.......(7)$
Now we will substitute the values of equation (5), (6) and (7) in $\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q)$ we get,
$\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q)=\left[ a+(p-1)\dfrac{d}{2} \right](q-r)+\left[ a+(q-1)\dfrac{d}{2} \right](r-p)+\left[ a+(r-1)\dfrac{d}{2} \right](p-q)$
$=a(q-r)+(p-1)\dfrac{d}{2}(q-r)+a(r-p)+(q-1)\dfrac{d}{2}(r-p)+a(p-q)+(r-1)\dfrac{d}{2}(p-q)$
Combining like terms together we get,
$=\left[ a(q-r)+a(r-p)+a(p-q) \right]+\left[ (p-1)\dfrac{d}{2}(q-r)+(q-1)\dfrac{d}{2}(r-p)+(r-1)\dfrac{d}{2}(p-q) \right]$
Taking $a$ common from first bracket and $\dfrac{d}{2}$ common from second bracket we get,
$=a\left[ q-r+r-p+p-q \right]+\dfrac{d}{2}\left[ (p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q) \right]$
$=a\left[ q-r+r-p+p-q \right]+\dfrac{d}{2}\left[ pq-pr-q+r+qr-qp-r+p+rp-rq-p+q \right]$
The first bracket becomes zero and the second bracket also becomes zero, so
$\begin{align}
  & =a\left[ 0 \right]+\dfrac{d}{2}\left[ 0 \right] \\
 & =0 \\
\end{align}$

Thus the value of $\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q)$ is zero.

Note: Sum of n terms in A.P is given by ${{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]$ ; where $a$ is the first term of A.P and d is the common difference. Care should be taken when we substitute the respective values of $\dfrac{x}{p},\dfrac{y}{q},\dfrac{z}{r}$ in$\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q)$ plus and minus sign should be written carefully.