Question

If the $p^{th},q^{th}$ and $r^{th}$ term of an arithmetic sequence are $a,b$ and $c$ respectively, then the value of $\left[a\right(q–r)+b(r–p)+c(p–q\left)\right]=$

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$0$

Explanation of the correct option.

Compute the required value.

Let $A$ be the first term of AP and $D$ be the common difference.

$p^{th}$term of A.P. $a=A-D+pD$………………..$\left(1\right)$

$q^{th}$ term of AP $b=A-D+qD$………………..$\left(2\right)$

$r^{th}$ term of AP $c=A-D+rD$………………...$\left(3\right)$

Multiply equation $\left(1\right)$,$\left(2\right)$ and $\left(3\right)$ by $\left(q-r\right),\left(r-p\right),\left(p-q\right)$ respectively and adding them together, we get

$\left[a\right(q–r)+b(r–p)+c(p–q\left)\right]=$$A\left(q-r\right)-D\left(q-r\right)+pD\left(q-r\right)+A(r-p)-D(r-p)+qD(r-p)+A(p-q)-D(p-q)+rD(p-q)$

$\Rightarrow$$\left[a\right(q–r)+b(r–p)+c(p–q\left)\right]=$$\left(A-D\right)\left[q-r+r-p+p-q\right]+D\left[pq-pr+qr-qp+rp-qr\right]$

$\Rightarrow$$\left[a\right(q–r)+b(r–p)+c(p–q\left)\right]=$$0$

Hence option $\left(C\right)$ is the correct option.