Question

If ${{p}^{th}}$, ${{q}^{th}}$ and ${{r}^{th}}$ terms of an A.P is a, b, c. Then find the value of $a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)$.
(a) 0
(b) 1
(c) $a+b+c$
(d) $abc$

Answer 1

Hint: We first write the general notation for ${{p}^{th}}$, ${{q}^{th}}$ and ${{r}^{th}}$ terms of A.P (Arithmetic Progression). We then equate general terms with the given values. After equating we multiply each term with the given multiples. We finally add all these terms to get the required answer.

Complete step-by-step solution:
We are given that the ${{p}^{th}}$, ${{q}^{th}}$ and ${{r}^{th}}$ terms of an A.P (Arithmetic Progression) is a, b, c. We need to find the value of $a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)$.
We know that ${{n}^{th}}$ term of an A.P is given as ${{T}_{n}}=a+\left( n-1 \right)d$, where ‘a’ is first term and ‘d’ is the common difference.
Using this we now write the ${{p}^{th}}$, ${{q}^{th}}$ and ${{r}^{th}}$ terms of A.P (Arithmetic Progression). Let us assume the first term be ‘f’ and the common difference is ‘g’.
According the problem we have ${{p}^{th}}$ term ${{T}_{p}}=a$, ${{q}^{th}}$ term ${{T}_{q}}=b$ and ${{r}^{th}}$ term ${{T}_{r}}=c$.
Now, ${{T}_{p}}=f+(p-1)g$.
$a=f+(p-1)g$ ---(1).
${{T}_{q}}=f+(q-1)g$.
$b=f+(q-1)g$ ---(2).
${{T}_{r}}=f+\left( r-1 \right)g$.
$c=f+\left( r-1 \right)g$ ---(3).
Let us multiply equation (1), with $\left( q-r \right)$ on both sides.
$a\left( q-r \right)=\left[ f+\left( p-1 \right)g \right]\times \left( q-r \right)$.
$a\left( q-r \right)=\left( \left( f-g \right)+pg \right)\times \left( q-r \right)$.
$a\left( q-r \right)=\left( \left( f-g \right)\times \left( q-r \right) \right)+\left( pg\times \left( q-r \right) \right)$.
$a\left( q-r \right)=\left( \left( f-g \right)\times \left( q-r \right) \right)+pqg-prg$ ---(4).
Let us multiply equation (2), with $\left( r-p \right)$ on both sides.
$b\left( r-p \right)=\left( f+\left( q-1 \right)g \right)\times \left( r-p \right)$.
$b\left( r-p \right)=\left( \left( f-g \right)+qg \right)\times \left( r-p \right)$.
$b\left( r-p \right)=\left( \left( f-g \right)\times \left( r-p \right) \right)+\left( qg\times \left( r-p \right) \right)$.
$b\left( r-p \right)=\left( \left( f-g \right)\times \left( r-p \right) \right)+qrg-pqg$ ---(5).
Let us multiply equation (3), with $(p-q)$ on both sides.
$c\left( p-q \right)=\left( f+\left( r-1 \right)g \right)\times \left( p-q \right)$.
$c\left( p-q \right)=\left( \left( f-g \right)+rg \right)\times \left( p-q \right)$.
$c\left( p-q \right)=\left( \left( f-g \right)\times \left( p-q \right) \right)+\left( rg\times \left( p-q \right) \right)$.
$c\left( p-q \right)=\left( \left( f-g \right)\times \left( p-q \right) \right)+prg-qrg$ ---(6).
We now add equations (4), (5) and (6).
$a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=\left( \left( f-g \right)\times \left( q-r \right) \right)+pqg-prg+\left( \left( f-g \right)\times \left( r-p \right) \right)+qrg-pqg+\left( \left( f-g \right)\times \left( p-q \right) \right)+prg-qrg$.
$a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=\left( \left( f-g \right)\times \left( \left( q-r \right)+\left( r-p \right)+\left( p-q \right) \right) \right)+pqg-prg++qrg-pqg++prg-qrg$.
$a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=\left( \left( f-g \right)\times \left( q-r+r-p+p-q \right) \right)+0$.
$a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=\left( f-g \right)\times 0$.
$a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=0$.
∴ The value of $a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)$ is 0.
The correct option is (a).

Note: We should not take ${{T}_{p}}=a+\left( p-1 \right)d$ as, variable ‘a’ is already used up to represent the value of ${{p}^{th}}$ term. If we take variable ‘a’ here, then the calculation will become wrong and leads to an unnecessary result. We always take the variables that are not already used up in the problem. We always need to start with the general term of given progression to solve such types of problems.