Question

If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are the $p^{th},q^{th},r^{th}$ terms respectively of an $A.P$ the the value of $ab(p-q)+bc(q-r)+ca(r-p)$ is

• A. $-1$
• B. $2$
• C. $0$
• D. $-2$

Answer 1

The correct option is C $0$

We know foe any $n^{th}$ term of an $A.P$, $n^{th}$ term $=a+(n-1)d$ where $a=$ first term and $d=$ common difference.

$\therefore a_1+(p-1)d=\frac{1}{a}$,

$a_1+(9-1)d=\frac{1}{h}$ and $a_1+(r-1)d=\frac{1}{c}$

$\therefore \frac{1}{a}=(a_1-d)+pd---\left(1\right),\frac{1}{b}=(a_1-d)+qd---\left(2\right)$ and $\frac{1}{c}=(a_1-d)+rd---\left(3\right)$

By $\left(1\right)-\left(2\right),\left(2\right)-\left(3\right)$ and $\left(3\right)-\left(1\right)$

$\frac{1}{a}-\frac{1}{b}=(p-q)d,\frac{1}{b},\frac{1}{c}=(q-r)d$ and

$\frac{1}{c}-\frac{1}{a}=(r-p)d$

$\therefore \frac{b-a}{d\left(ab\right)}=p-q,\frac{c-b}{d\left(bc\right)}=q-r,\frac{a-c}{d\left(ac\right)}=r-p$

or, $\frac{b-a}{d}=ab(p-q),\frac{c-b}{d}=bc(q-r),\frac{a-c}{d}=ac(r-p)$

Adding all the equation

$\frac{(b-a)+(c-b)+(a-c)}{d}=ab(p-q)+bc(q-r)+ac(r-p)$

or, $ab(p-q)+bc(q-r)+ac(r-p)=0$------Proved,