Question

Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that
 $ \dfrac{a}{p}(q - r) + \dfrac{b}{q}(r - p) + \dfrac{c}{r}(p - q) = 0 $

Answer 1

Hint: Here we will use the standard formula for the Arithmetic progression and will follow the given three conditions for the three A.P. and then simplify as per the given LHS of the equation. Use summation of “n” terms formula.

Complete step-by-step answer:
Sum of “n” terms in the Arithmetic Progression can be given by –
 $ {S_n} = \dfrac{n}{2}[2A + (n - 1)D] $
Where, $ {S_n} = $ Sum of first “n” terms
 $ A = $ First term
 $ D = $ Common difference between the two consecutive terms
Given that sum of first “P” terms is “a”
Place the given values in the standard formula-
 $ \Rightarrow a = \dfrac{p}{2}[2A + (p - 1)D]{\text{ }} $
When any term in the multiplicative on one side changes its side then it goes to the denominator.
 $ \Rightarrow \dfrac{a}{p} = \dfrac{1}{2}[2A + (p - 1)D]{\text{ }}....{\text{(i) }} $
Similarly place values for the condition –
The sum of first “q” terms is “b”
 $ \Rightarrow b = \dfrac{q}{2}[2A + (q - 1)D]{\text{ }} $
When any term in the multiplicative on one side changes its side then it goes to the denominator.
 $ \Rightarrow \dfrac{b}{q} = \dfrac{1}{2}[2A + (q - 1)D]{\text{ }}....{\text{(ii)}} $
The sum of first “r” terms is “c”
 $ \Rightarrow c = \dfrac{r}{2}[2A + (r - 1)D]{\text{ }} $
When any term in the multiplicative on one side changes its side then it goes to the denominator.
 $ \Rightarrow \dfrac{c}{r} = \dfrac{1}{2}[2A + (r - 1)D]{\text{ }}......{\text{ (iii)}} $
As per the required left hand side of the equation –
Multiply equation (i) by $ (q - r) $
 $ \Rightarrow \dfrac{a}{p}(q - r) = \dfrac{1}{2}[2A + (p - 1)D](q - r){\text{ }} $
Simplify left hand side of the equation –
 $ \Rightarrow \dfrac{a}{p}(q - r) = [A + \dfrac{{(p - 1)}}{2}D](q - r){\text{ }}....{\text{(iv)}} $
Multiply equation (ii) by $ (r - p) $
 $ \Rightarrow \dfrac{b}{q}(q - r) = [A + \dfrac{{(q - 1)}}{2}D](r - p){\text{ }}....{\text{(v)}} $
Multiply equation (iii) by $ (p - q) $
 $ \Rightarrow \dfrac{c}{r}(p - q) = [A + \dfrac{{(r - 1)}}{2}D](p - q){\text{ }}....{\text{(vi)}} $
Add RHS and LHS of the equations - $ (iv),(v){\text{ and (vi)}} $ Take common factor where applicable
 $
   \Rightarrow \dfrac{a}{p}(q - r) + \dfrac{b}{q}(r - p) + \dfrac{c}{r}(p - q) = A(q - r + r - p + p - q) + \\
  \dfrac{D}{2}[pq - pr - q + rq - pq - r + p + rp - rq - p + q] \;
  $
Same terms with opposite signs cancel each other in the above equation.
 $ \Rightarrow \dfrac{a}{p}(q - r) + \dfrac{b}{q}(r - p) + \dfrac{c}{r}(p - q) = A(0) + \dfrac{D}{2}[0] $
Simplify – Anything multiplied with zero gives resultant zero.
 $ \Rightarrow \dfrac{a}{p}(q - r) + \dfrac{b}{q}(r - p) + \dfrac{c}{r}(p - q) = 0 $
Hence the required equation is proved.

Note: Always remember the difference between the nth term and summation of “n” terms and use the formula accordingly. Also, refer to geometric progression (G.P.) terms and remember the difference between A.P and G.P.