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If-a-x-px-a-y-qy-a-z-rz-and-p-q-r-are-in-AP-then-x-y-z-will-be-in-




Question Number 46924 by 786786AM last updated on 02/Nov/18
If  ((a−x)/(px)) = ((a−y)/(qy)) = ((a−z)/(rz)) and p, q, r  are  in AP, then x, y, z  will be in
$$\mathrm{If}\:\:\frac{{a}−{x}}{{px}}\:=\:\frac{{a}−{y}}{{qy}}\:=\:\frac{{a}−{z}}{{rz}}\:\mathrm{and}\:{p},\:{q},\:{r}\:\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{AP},\:\mathrm{then}\:{x},\:{y},\:{z}\:\:\mathrm{will}\:\mathrm{be}\:\mathrm{in} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18
((a−x)/(px))=((a−y)/(qy))=((a−z)/(rz))=(1/k)(say)  px=k(a−x)  p=((k(a−x))/x)  q==((k(a−y))/y)  r=((k(a−z))/z)  2q=p+r  ((2k(a−y))/y)=((k(a−x))/x)+((k(a−z))/z)  ((2(a−y))/y)=((a−x)/x)+((a−z)/z)  ((2a)/y)−2=(a/x)−1+(a/z)−1  (2/y)=(1/x)+(1/z)  so x,y,z  are in HP
$$\frac{{a}−{x}}{{px}}=\frac{{a}−{y}}{{qy}}=\frac{{a}−{z}}{{rz}}=\frac{\mathrm{1}}{{k}}\left({say}\right) \\ $$$${px}={k}\left({a}−{x}\right) \\ $$$${p}=\frac{{k}\left({a}−{x}\right)}{{x}} \\ $$$${q}==\frac{{k}\left({a}−{y}\right)}{{y}} \\ $$$${r}=\frac{{k}\left({a}−{z}\right)}{{z}} \\ $$$$\mathrm{2}{q}={p}+{r} \\ $$$$\frac{\mathrm{2}{k}\left({a}−{y}\right)}{{y}}=\frac{{k}\left({a}−{x}\right)}{{x}}+\frac{{k}\left({a}−{z}\right)}{{z}} \\ $$$$\frac{\mathrm{2}\left({a}−{y}\right)}{{y}}=\frac{{a}−{x}}{{x}}+\frac{{a}−{z}}{{z}} \\ $$$$\frac{\mathrm{2}{a}}{{y}}−\mathrm{2}=\frac{{a}}{{x}}−\mathrm{1}+\frac{{a}}{{z}}−\mathrm{1} \\ $$$$\frac{\mathrm{2}}{{y}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}\:\:{so}\:{x},{y},{z}\:\:{are}\:{in}\:{HP} \\ $$

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