Question

Question 11 Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that a(q-r) / p + b(r-p) / q + c(p-q) / r = 0

Class X1 - Maths -Sequences and Series Page 185

Answer 1

Let the first term is A and common difference is d .
Sp = a
P/2{2A + (P-1)d} = a-----(1)

Sq = b
q/2{2A + (q-1)d} = b------(2)

Sr = c
r/2{2A + (r -1)d} = c------(3)

Now,
LHS = a(q-r)/p + b(r-p)/q + c(p-q)/r
= (q-r)/p ×P/2{2A + (p-1)d} + (r-p)/q ×q/2{2A + (q -1)d} + (p-q)/r × r/2{2A + (r-1)d}

= 1/2 [(q-r)(2A +pd -d) + (r -p)(2A + qd - d)+(p-q)(2A + rd-d)]

= 1/2[2A{(q-r)+(r-p)+(p-q)} +d{p(q-r)+q(r-p)+r(p-q)-d{(q-r)+(r-p)+(p-q)}]

= 1/2[2A{0} +d{pq-pr+qr-qp+rp-rq}-d{0}]

=1/2[0 + 0 + 0]

= 0 = RHS

Answer 2

Let  the first  term of AP be  A  and common difference  be  d,

We  know that sum if  n terms  of  AP,

$S_n=\frac{n}{2}[2a+(n-1)d]$

Using  this  formula,

Sum of  p term of AP,

$S_p=\frac{p}{2}[2A+(p-1)d]\\\;\\a=\frac{p}{2}[2A+(p-1)d]\\\;\\\frac{a}{p}=\frac{1}{2}[2A+pd-d]\\\;\\\frac{a}{p}=\frac{1}{2}[(2A-d)+pd]\\\;\\\frac{a}{p}(q-r)=\frac{1}{2}[(2A-d)+pd](q-r)\\\;\\\frac{a}{p}(q-r)=\frac{1}{2}[(q-r)(2A-d)+(q-r)pd].........i)$

Sum of  q terms

$S_q=\frac{q}{2}[2A+(q-1)d]\\\;\\b=\frac{q}{2}[2A+(q-1)d]\\\;\\\frac{b}{q}=\frac{1}{2}[2A+qd-d]\\\;\\\frac{b}{q}=\frac{1}{2}[(2A-d)+qd]\\\;\\\frac{b}{q}(r-p)=\frac{1}{2}[(2A-d)+qd](r-p)\\\;\\\frac{b}{q}(r-p)=\frac{1}{2}[(r-p)(2A-d)+(r-p)qd].........ii)$

Similarly,

for Sum of  r term of  AP,

$\frac{c}{r}(p-q)=\frac{1}{2}[(p-q)(2A-d)+(p-q)rd]........iii)$

Adding  eq. i) , ii) and  iii),

$\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=\frac{1}{2}[(q-r)(2A-d)+(q-r)pd]+\frac{1}{2}[(r-p)(2A-d)+(r-p)qd]+\frac{1}{2}[(p-q)(2A-d)+(p-q)rd]\\\;\\\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=\frac{1}{2}[(2A-d)((q-r)+(r-p)+(p-q)]+\frac{1}{2}[(q-r)pd+(r-p)qd+(p-q)rd]\\\;\\\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=\frac{1}{2}[(2A-d)(q-r+r-p+p-q)]+\frac{1}{2}[qpd-rpd+rqd-pqd+rpd-rqd]\\\;\\\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=\frac{1}{2}[(2A-d)(0)]+\frac{1}{2}[0]$

$\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0+0\\\;\\\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$