Question

If a, b and c be respective sum of first p, q ans r terms of A.P. Show that:
$\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

Answer 1

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$

Answer 2

Answer:

Hope it will help uh .....