Question

If $a,b,c$ are $p^{th},q^{th},r^{th}$ terms respectively of an H.P.then $ab(p-q)+bc(q-r)+ca(r-p)$ equals

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

Answer: (B)

$0$

Given a,b,c are in H.P
$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P
let first terms of A.P. is A and common difference is d
$\therefore \frac{1}{a}=A+(p-1)d$
$\frac{1}{b}=A+(q-1)d$
$\frac{1}{c}=A+(r-1)d$
$bc(q-r)=abc[A+(p-1\left)d\right](q-r)$ ....(1)
$ca(r-p)=abc[A+(q-1\left)d\right](r-p)$ ....(2)
$ab(p-q)=abc[A+(r-1\left)d\right](p-q)$ ....(3)

on adding (1), (2) & (3), we get
$bc(q-r)+ca(r-p)+ab(p-q)=abc\{A(q-r+r-p+p-q)+d[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]\}$
Therefore, $bc(q-r)+ca(r-p)+ab(p-q)=0$
Ans: B