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Question: If \({{p}^{th}},{{q}^{th}}\text{ and }{{r}^{th}}\) term of an A.P. are a , b and c, respectively. Th...

If pth,qth and rth{{p}^{th}},{{q}^{th}}\text{ and }{{r}^{th}} term of an A.P. are a , b and c, respectively. Then show that (q-r)a+(r-p)b+(p-q)c=0.

Explanation

Solution

Hint: Try to get a relation between p, q, r and a, b, c, respectively using the formula of general term of an A.P. Now substitute the value of p, q and r using the equations in the left-hand side of the equation given in the question and solve to get the required result.

Complete step-by-step answer:
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by Tn{{T}_{n}}, and sum till n terms is denoted by Sn{{S}_{n}} .
Tn=a1+(n−1)d{{T}_{n}}={{a}_{1}}+\left( n-1 \right)d
Sn=n2(2a1+(n−1)d)=n2(a1+an){{S}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)=\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)
Now moving to the solution of the question, it is given that pth,qth and rth{{p}^{th}},{{q}^{th}}\text{ and }{{r}^{th}} term of an A.P. are a , b and c. Therefore, using the formula of the general term of an A.P., we can say
a=a1+(p−1)da={{a}_{1}}+(p-1)d
⇒a−a1d+1=p..............(i)\Rightarrow \dfrac{a-{{a}_{1}}}{d}+1=p..............(i)
b=a1+(q−1)db={{a}_{1}}+(q-1)d
⇒b−a1d+1=q..............(ii)\Rightarrow \dfrac{b-{{a}_{1}}}{d}+1=q..............(ii)

c=a1+(r−1)dc={{a}_{1}}+(r-1)d
⇒c−a1d+1=r..............(iii)\Rightarrow \dfrac{c-{{a}_{1}}}{d}+1=r..............(iii)

Now if we substitute the values of p, q and r in the left-hand side of the equation given in the question, we get
(q−r)a+(r−p)b+(p−q)c\left( q-r \right)a+\left( r-p \right)b+\left( p-q \right)c
=(b−a1d+1−(c−a1d+1))a+(c−a1d+1−(a−a1d+1))b+(a−a1d+1−(b−a1d+1))c=\left( \dfrac{b-{{a}_{1}}}{d}+1-\left( \dfrac{c-{{a}_{1}}}{d}+1 \right) \right)a+\left( \dfrac{c-{{a}_{1}}}{d}+1-\left( \dfrac{a-{{a}_{1}}}{d}+1 \right) \right)b+\left( \dfrac{a-{{a}_{1}}}{d}+1-\left( \dfrac{b-{{a}_{1}}}{d}+1 \right) \right)c
=(b−a1−c+a1d+1−1)a+(c−a1−a+a1d+1−1)b+(a−a1−b+a1d+1−1)c=\left( \dfrac{b-{{a}_{1}}-c+{{a}_{1}}}{d}+1-1 \right)a+\left( \dfrac{c-{{a}_{1}}-a+{{a}_{1}}}{d}+1-1 \right)b+\left( \dfrac{a-{{a}_{1}}-b+{{a}_{1}}}{d}+1-1 \right)c
=(b−cd)a+(c−ad)b+(a−bd)c=\left( \dfrac{b-c}{d} \right)a+\left( \dfrac{c-a}{d} \right)b+\left( \dfrac{a-b}{d} \right)c
Now if we multiply and open the brackets to simplify the above expression, we get
ab−ac+cb−ab+ac−bcd=0\dfrac{ab-ac+cb-ab+ac-bc}{d}=0
So, as we have shown that left-hand side of the equation given in the question is equal to zero, we can say that we have proved the result (q−r)a+(r−p)b+(p−q)c=0\left( q-r \right)a+\left( r-p \right)b+\left( p-q \right)c=0 .

Note: It is not always necessary that you can notice that the sequences are in the form of simple arithmetic progressions and draw results, but it is for sure that the terms of a sequence will have some order, and it depends on you how wisely you figure out the pattern. The suggested approach can be to try to find the general term of the sequence by observing the pattern of numbers appearing in the sequence and proceed.