Question Number 11979 by 786786AM last updated on 08/Apr/17
$$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:{p}\:\mathrm{terms},\:\mathrm{first}\:\:{q}\:\mathrm{terms}\:\mathrm{and} \\ $$$$\mathrm{first}\:{r}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{be}\:\:{x},\:{y}\:\:\mathrm{and}\:{z}\: \\ $$$$\mathrm{respectively}.\:\mathrm{Then} \\ $$$$\frac{{x}}{{p}}\left({q}−{r}\right)\:+\:\frac{{y}}{{q}}\left({r}−{p}\right)\:+\:\frac{{z}}{{r}}\left({p}−{q}\right)\:\:\mathrm{is} \\ $$
Answered by ajfour last updated on 09/Apr/17
![(x/p)=a+(p−1)(d/2) (y/q)=a+(q−1)(d/2) (z/r)=a+(r−1)(d/2) so, (x/p)(q−r)+(y/q)(r−p)+(z/r)(p−q) =( a−(d/2))(q−r+r−p+p−q) +(d/2)[p(q−r)+q(r−p)+r(p−q)] =(a−(d/2))(0) + (d/2)(0) =0 .](https://www.tinkutara.com/question/Q11983.png)
$$\frac{{x}}{{p}}={a}+\left({p}−\mathrm{1}\right)\frac{{d}}{\mathrm{2}} \\ $$$$\frac{{y}}{{q}}={a}+\left({q}−\mathrm{1}\right)\frac{{d}}{\mathrm{2}} \\ $$$$\frac{{z}}{{r}}={a}+\left({r}−\mathrm{1}\right)\frac{{d}}{\mathrm{2}} \\ $$$${so},\:\frac{{x}}{{p}}\left({q}−{r}\right)+\frac{{y}}{{q}}\left({r}−{p}\right)+\frac{{z}}{{r}}\left({p}−{q}\right) \\ $$$$\:\:\:=\left(\:{a}−\frac{{d}}{\mathrm{2}}\right)\left({q}−{r}+{r}−{p}+{p}−{q}\right) \\ $$$$\:\:\:\:\:+\frac{{d}}{\mathrm{2}}\left[{p}\left({q}−{r}\right)+{q}\left({r}−{p}\right)+{r}\left({p}−{q}\right)\right] \\ $$$$\:\:\:\:=\left({a}−\frac{{d}}{\mathrm{2}}\right)\left(\mathrm{0}\right)\:+\:\frac{{d}}{\mathrm{2}}\left(\mathrm{0}\right) \\ $$$$\:\:\:=\mathrm{0}\:. \\ $$