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Question Number 11979 by 786786AM last updated on 08/Apr/17

$$\mathrm{If}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{first}p\mathrm{terms},\mathrm{first}q\mathrm{terms}\mathrm{and}$$$$\mathrm{first}r\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{AP}\mathrm{be}x,y\mathrm{and}z$$$$\mathrm{respectively}.\mathrm{Then}$$$$\frac{x}{p}(q-r)+\frac{y}{q}(r-p)+\frac{z}{r}(p-q)\mathrm{is}$$

Answered by ajfour last updated on 09/Apr/17

$$\frac{x}{p}=a+(p-1)\frac{d}{2}$$$$\frac{y}{q}=a+(q-1)\frac{d}{2}$$$$\frac{z}{r}=a+(r-1)\frac{d}{2}$$$$so,\frac{x}{p}(q-r)+\frac{y}{q}(r-p)+\frac{z}{r}(p-q)$$$$=(a-\frac{d}{2})(q-r+r-p+p-q)$$$$+\frac{d}{2}[p(q-r)+q(r-p)+r(p-q)]$$$$=(a-\frac{d}{2})\left(0\right)+\frac{d}{2}\left(0\right)$$$$=0.$$

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