Browse Questions

# If $p^{th},\:q^{th}\:\:and\:\:r^{th}$ terms oa an A.P. are $a,\:b\:and\:c$ respectively, then show that $(q-r)a+(r-p)b+(p-q)c=0$

Given $p^{th}$ term of an A.P.$a$, $q^{th}$ term $=b$ and $r^{th}$ term $=c$
$\Rightarrow\:t_p=A+(p-1)D=a$ where $A$ is first term and $D$ is common difference
$t_q=A+(q-1)D=b$ and $A+(r-1)D=c$
Now
$(q-r)a=(q-r).\big[A+(p-1)D\big]$
$\qquad\:=A(q-r)+(p-1)(q-r)D$
$\qquad\:=A(q-r)+D(pq-pr-q+r)$
Step 2
$(r-p)b=(r-p).\big[A+(q-1)D\big]$
$\qquad\:=A(r-p)+(q-1)(r-p)D\big]$
$\qquad\:=A(r-p)+D(qr-pq-r+p)$
Step 3
Similarly $(p-q)c=(p-q).\big[A+(r-1)D\big]$
$\qquad\:=A(p-q)+(r-1)(p-q)D\big]$
$\qquad\:=A(p-q)+D(pr-qr-p+q)$
Step 4
Adding the three we get $(q-r)a+(r-p)b+(p-q)c=$
$A(q-r)+D(pq-pr-q+r)+A(r-p)+D(qr-pq-r+p)+A(p-q)+D(pr-qr-p+q)$
$=A(0)+D(0)=0$
Hence proved.