Question

If the $p^{th},q^{th},r^{th}$ term of an A.P. be $a,b,c$ respectively then show that $a(q-r)+b(r-p)+c(p-q)=0$.

Answer 1

$n^{th}$ term of an A.P is $a_n=a_1+(n-1)d$

where, $a_1$ is first term and $d$ is the common difference.

$\therefore r^{th}$ term $\Rightarrow a_r=c=a_1+(r-1)d$

$\therefore q^{th}$ term $\Rightarrow a_q=b=a_1+(q-1)d$

$\therefore p^{th}$ term $\Rightarrow a_p=a=a_1+(p-1)d$

$\therefore a(q-r)+b(r-p)+c(p-q)=[a_1+(p-1\left)d\right](q-r)+[a_1+(q-1\left)d\right](r-p)+[a_1+(r-1\left)d\right](p-q)$

$⟹a_1(q-r+r-p+p-q)+d\left[\right(q-r\left)\right(p-1)+(q-1\left)\right(r-p)+(r-1\left)\right(p-q\left)\right]$

$⟹a_1\left(0\right)+d[qp-pr+r-q+qr+p-qp-r+rp+q-p-rq)$

$⟹a_1\left(0\right)+d\left(0\right)=0$

$\therefore a(q-r)+b(r-p)+c(p-q)=0$