Question

If we are given the terms \[{{p}^{th}},{{q}^{th}},{{r}^{th}}\] of an AP as a, b and c, then prove that \[a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=0.\]

Answer 1

Hint: In this question first of all consider an AP with the first term as A and common difference as D. Now, use the formula for the general term that is \[{{a}_{n}}=a+\left( n-1 \right)d\] to write the \[{{p}^{th}},{{q}^{th}},{{r}^{th}}\] terms of an AP and substitute the values of a, b and c in LHS of the given result to prove the required result.

Complete step-by-step solution:
In this question, we are given that \[{{p}^{th}},{{q}^{th}},{{r}^{th}}\] terms of an AP are a, b and c. Then we have to prove that \[a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=0.\] Before proceeding with the question, let us see what an AP is.
An Arithmetic Progression (AP) is basically a sequence of numbers such that the difference between the consecutive terms is constant. For example, 2, 4, 6, 8 is an arithmetic progression with a common difference of 2.
The general term or the \[{{n}^{th}}\] term of an AP or Arithmetic Progression is given by
\[{{a}_{n}}=a+\left( n-1 \right)d......\left( i \right)\]
where ‘a’ and ‘d’ are the first term and common difference of an AP respectively.
Let us now consider our question. Let us take an AP with the first term as A and the common difference as D. Now, we are given that the \[{{p}^{th}}\] term of this AP is ‘a’. So, by using the general term formula in equation (i), we get,
\[a=A+\left( p-1 \right)D......\left( ii \right)\]
Similarly, we are given that \[{{q}^{th}}\] term of this AP is b. So, again using the general term formula in equation (i), we get,
\[b=A+\left( q-1 \right)D......\left( iii \right)\]
Similarly, we are given that \[{{r}^{th}}\] term of this AP is c. So, again using the general term formula in equation (i), we get,
\[c=A+\left( r-1 \right)D......\left( iv \right)\]
We have to prove that \[a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=0.\] Let us take the LHS of this result.
\[LHS=a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)\]
By substituting the values of a, b and c from equation (ii), (iii) and (iv) respectively in the above expression, we get,
\[\Rightarrow LHS=\left[ A+\left( p-1 \right)D \right]\left( q-r \right)+\left[ A+\left( q-1 \right)D \right]\left( r-p \right)+\left[ A+\left( r-1 \right)D \right]\left( p-q \right)\]
By simplifying the above expression, we get,
\[\begin{align}
  & \Rightarrow LHS=Aq-Ar+\left( p-1 \right)Dq-\left( p-1 \right)Dr+Ar-Ap+\left( q-1 \right)Dr-\left( q-1 \right)Dp \\
 & +Ap-Aq+\left( r-1 \right)Dp-\left( r-1 \right)Dq \\
\end{align}\]
By rearranging the terms of the above expression and further simplifying it, we get,
\[\begin{align}
  & \Rightarrow LHS=\left( Aq-Ar+Ar-Ap+Ap-Aq \right)+ \\
 & \left( pDq-Dq-pDr+Dr+qDr-Dr-qDp+Dp+rDp-Dp-rDq+Dq \right) \\
\end{align}\]
By cancelling the like terms, we finally get,
\[\Rightarrow LHS=0\]
\[\Rightarrow LHS=RHS\]
Hence, we have proved that \[a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=0.\]

Note: Many students make this mistake of taking the first term as ‘a’ instead of ‘A’ which is wrong because we are given the question that ‘a’ is \[{{p}^{th}}\] term of AP. So, if we also take ‘a’ as the first term of AP, that will create lots of confusion in the question. So, students must remember to assume the extra variable different from those already given in the question.