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Question: If \({{p}^{th}}\), \({{q}^{th}}\) and \({{r}^{th}}\) terms of an A.P is a, b, c. Then find the value...

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

Explanation

Solution

We first write the general notation for pth{{p}^{th}}, qth{{q}^{th}} and rth{{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 pth{{p}^{th}}, qth{{q}^{th}} and rth{{r}^{th}} terms of an A.P (Arithmetic Progression) is a, b, c. We need to find the value of a(q−r)+b(r−p)+c(p−q)a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right).
We know that nth{{n}^{th}} term of an A.P is given as Tn=a+(n−1)d{{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 pth{{p}^{th}}, qth{{q}^{th}} and rth{{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 pth{{p}^{th}} term Tp=a{{T}_{p}}=a, qth{{q}^{th}} term Tq=b{{T}_{q}}=b and rth{{r}^{th}} term Tr=c{{T}_{r}}=c.
Now, Tp=f+(p−1)g{{T}_{p}}=f+(p-1)g.
a=f+(p−1)ga=f+(p-1)g ---(1).
Tq=f+(q−1)g{{T}_{q}}=f+(q-1)g.
b=f+(q−1)gb=f+(q-1)g ---(2).
Tr=f+(r−1)g{{T}_{r}}=f+\left( r-1 \right)g.
c=f+(r−1)gc=f+\left( r-1 \right)g ---(3).
Let us multiply equation (1), with (q−r)\left( q-r \right) on both sides.
a(q−r)=[f+(p−1)g]×(q−r)a\left( q-r \right)=\left[ f+\left( p-1 \right)g \right]\times \left( q-r \right).
a(q−r)=((f−g)+pg)×(q−r)a\left( q-r \right)=\left( \left( f-g \right)+pg \right)\times \left( q-r \right).
a(q−r)=((f−g)×(q−r))+(pg×(q−r))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(q−r)=((f−g)×(q−r))+pqg−prga\left( q-r \right)=\left( \left( f-g \right)\times \left( q-r \right) \right)+pqg-prg ---(4).
Let us multiply equation (2), with (r−p)\left( r-p \right) on both sides.
b(r−p)=(f+(q−1)g)×(r−p)b\left( r-p \right)=\left( f+\left( q-1 \right)g \right)\times \left( r-p \right).
b(r−p)=((f−g)+qg)×(r−p)b\left( r-p \right)=\left( \left( f-g \right)+qg \right)\times \left( r-p \right).
b(r−p)=((f−g)×(r−p))+(qg×(r−p))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(r−p)=((f−g)×(r−p))+qrg−pqgb\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)(p-q) on both sides.
c(p−q)=(f+(r−1)g)×(p−q)c\left( p-q \right)=\left( f+\left( r-1 \right)g \right)\times \left( p-q \right).
c(p−q)=((f−g)+rg)×(p−q)c\left( p-q \right)=\left( \left( f-g \right)+rg \right)\times \left( p-q \right).
c(p−q)=((f−g)×(p−q))+(rg×(p−q))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(p−q)=((f−g)×(p−q))+prg−qrgc\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(q−r)+b(r−p)+c(p−q)=((f−g)×(q−r))+pqg−prg+((f−g)×(r−p))+qrg−pqg+((f−g)×(p−q))+prg−qrga\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(q−r)+b(r−p)+c(p−q)=((f−g)×((q−r)+(r−p)+(p−q)))+pqg−prg++qrg−pqg++prg−qrga\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(q−r)+b(r−p)+c(p−q)=((f−g)×(q−r+r−p+p−q))+0a\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(q−r)+b(r−p)+c(p−q)=(f−g)×0a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=\left( f-g \right)\times 0.
a(q−r)+b(r−p)+c(p−q)=0a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=0.
∴ The value of a(q−r)+b(r−p)+c(p−q)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 Tp=a+(p−1)d{{T}_{p}}=a+\left( p-1 \right)d as, variable ‘a’ is already used up to represent the value of pth{{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.