Question

If the $ {p^{th}},{q^{th}}{\text{ and }}{r^{th}} $ term of an A.P. are a, b and c respectively. Prove that $ a\left( {q - r} \right) + b\left( {r - p} \right) + c\left( {p - q} \right) = 0 $ .

Answer 1

Hint: In this question, we need to prove that the relation $ a\left( {q - r} \right) + b\left( {r - p} \right) + c\left( {p - q} \right) = 0 $ holds true such that the $ {p^{th}},{q^{th}}{\text{ and }}{r^{th}} $ term of an A.P. are a, b and c. For this, we will use the relation between the $ {n^{th}} $ term of an AP series with the first term and the common difference of the series and try to relate the things according to the demand of the problem.

Complete step-by-step answer:
Let the first term of the Arithmetic Progression be ‘A’ and the common difference be ‘D’.
The $ {n^{th}} $ term of an AP is given as: $ {t_n} = a + \left( {n - 1} \right)d $ where, ‘a’ is the first term and ‘d’ is the common difference of the AP.
According to the question, the $ {p^{th}} $ term of the AP is a. So, we can write
  $ a = A + \left( {p - 1} \right)D - - - - (i) $
Also, it has been given that the $ {q^{th}} $ term of the AP is b. So, we can write
  $ b = A + \left( {q - 1} \right)D - - - - (ii) $
Again, the $ {r^{th}} $ term of the AP is c. So, we can write
  $ c = A + \left( {r - 1} \right)D - - - - (iii) $
Subtracting equation (ii) from (i), we get
 $
\Rightarrow a - b = A + (p - 1)D - A - (q - 1)D \\
    = (p - q)D - - - - (iv) \;
  $
Similarly, subtracting equation (iii) from (ii), we get
 $
\Rightarrow b - c = A + (q - 1)D - A - (r - 1)D \\
    = (q - r)D - - - - (v) \;
  $
Again, subtracting equation (iii) from (i), we get
 $
\Rightarrow c - a = A + (r - 1)D - A - (p - 1)D \\
    = (r - p)D - - - - (vi) \;
  $
Now multiplying equation (iv) by ‘c’, we get
  $ c\left( {a - b} \right) = c(p - q)D - - - - (vii) $
Also, multiplying equation (v) by ‘a’, we get
  $ a\left( {b - c} \right) = a(q - r)D - - - - (viii) $
Similarly, multiplying equation (vi) by ‘b’, we get
  $ b\left( {c - a} \right) = b(r - p)D - - - - (ix) $
Now, adding equation (vii), (viii) and (ix), we get
 $
\Rightarrow c(a - b) + a(b - c) + b(c - a) = c(p - q)D + a(q - r)D + b(q - r)D \\
\Rightarrow c(p - q) + a(q - r) + b(q - r) = ca - cb + ab - ac + bc - ab \\
\Rightarrow c(p - q) + a(q - r) + b(q - r) = 0 \;
  $
Hence, the given expression $ a\left( {q - r} \right) + b\left( {r - p} \right) + c\left( {p - q} \right) = 0 $ is true.

Note: In these types of questions, we need to carry out our calculations according to the demand of the problem. Here, we need to multiply the terms a, b and c as per the given expression. It is interesting to note that the first term and the common difference have neither been given in the question nor it is available in the expression to be proved.