Question

If $p^{th},q^{th},r^{th}$ terms of an A.P are $a,b,c$, then $a(q-r)+b(r-p)+c(p-q)=$

• A. $0$
• B. $1$
• C. $a+b+c$
• D. $abc$

Answer 1

Answer: (A)

$0$

We know that term number n of an A.P is
$a_n=a+(n-1)d$
Hence, $a=a_1+(p-1)d$ ...(i)
Similarly $b=a_1+(q-1)d$ ...(ii)
$c=a_1+(r-1)d$ ...(iii)
Subtracting equation (ii) from (i) gives us
$a-b=(p-q)d$
$d=\frac{a-b}{p-q}$
Substituting in (i) gives us
$a=a_1+\frac{(p-1)(a-b)}{p-q}$
$a_1=a-\frac{(p-1)(a-b)}{p-q}$
Substituting in (iii), we get
$c=a-\frac{(p-1)(a-b)}{p-q}+\frac{(r-1)(a-b)}{p-q}$
$a-c=\frac{(p-1)(a-b)}{p-q}-\frac{(r-1)(a-b)}{p-q}$
$a-c=\frac{a-b}{p-q}(p-r)$
$(a-c)(p-q)=(a-b)(p-r)$
$a(p-q)=c(p-q)=a(p-r)-b(p-r)$
$a(p-q-(p-r\left)\right)-c(p-q)+b(p-r)=0$
$a(r-q)+b(p-r)-c(p-q)=0$
$a(r-q)+b(p-r)+c(q-p)=0$