Question

If $p^{th}$, $q^{th}$ and $r^{th}$ terms of an A.P. are a, b, c respectively, then show that: $a(q–r)+b(r–p)+c(p–q)=0$

Answer 1

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

a = $p^{th}$ term

a = A + (p – 1) D .... (i)

b = $q^{th}$ term

b = A + (q – 1) D .... (ii)

c = $r^{th}$ term

c = A+ (r – 1) D .... (iii)

We have,

$a(q–r)+b(r–p)+c(p–q)$

$=A+(p–1)D(q–r)+A+(q–1)(r–p)+A+(r–1)D(p–q)$ [Using equations (i), (ii) and (iii)]

$=A(q–r)+(r–p)+(p–q)+D(p–1)(q–r)+(q–1)(r–p)+(r–1)(p–q)$

$=A(q–r)+(r–p)+(p–q)+D(p–1)(q–r)+(q–1)(r–p)+(r–1)(p–q)$

$=A.0+Dp(q–r)+q(r–p)+r(p–q)–(q–r)–(r–p)–(p–q)$

$=A.0+D.0=0$