Question

If p, q, r are the ath, bth, cth terms of an A.P then p(b - c) + q(c - a) + r(a - b) =

Answer 1

Answer:

Let A be the first term and D the common difference of A.P.

Tp=a=A+(p−1)D=(A−D)+pD (1)

Tq =b=A+(q−1)D=(A−D)+qD ..(2)

Tr =c=A+(r−1)D=(A−D)+rD ..(3)

Here we have got two unknowns A and D which are to be eliminated.

We multiply (1),(2) and (3) by q−r,r−p and p−q respectively and add:

a(q−r)+b(r−p)+c(p−q)

=(A−D)[q−r+r−p+p−q]+D[p(q−r)+q(r−p)+r(p−q)]

=0.

Answer 2

Given,

Here p,q and r are the$\[{a^{th}}\]$, $\[{b^{th}}\]$ and $\[{c^{th}}\]$term of an A.P.

Solution,

Consider an A.P. whose first term is A and the common difference is d.

Therefore,

$\[ \Rightarrow p = A + \left( {a - 1} \right)d\]$

$\[ \Rightarrow q = A + \left( {b - 1} \right)d\]$

$\[ \Rightarrow r = A + \left( {c - 1} \right)d\]$

Apply the above value to the equation $\[p\left( {b - c} \right) + q\left( {c - a} \right) + r\left( {a - b} \right).\]$

Therefore,

$\[ \Rightarrow \left[ {\left( {A + \left( {a - 1} \right)d} \right)\left( {b - c} \right)} \right] + \left[ {\left( {A + \left( {b - 1} \right)d} \right)\left( {c - a} \right)} \right] + \left[ {\left( {A + \left( {c - 1} \right)d} \right)\left( {a - b} \right)} \right]\]$

$\[ \Rightarrow \left[ {\left( {A + ad - d} \right)\left( {b - c} \right)} \right] + \left[ {\left( {A + bd - d} \right)\left( {c - a} \right)} \right] + \left[ {\left( {A + cd - d} \right)\left( {a - b} \right)} \right]\]$

$\[ \Rightarrow \left[ {Ab - Ac + abd - acd + dc} \right] + \left[ {Ac - Aa + bcd - abd + ad} \right] + \left[ {Aa - Ab + acd - bcd - ad + bd} \right]\]$$\[ \Rightarrow 0\]$

Hence, the value  $\[p\left( {b - c} \right) + q\left( {c - a} \right) + r\left( {a - b} \right)\]$ is$\[0.\]$