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Question Number 7526 by Little last updated on 02/Sep/16
If a, b, c are in AP then (a/(bc)) , (1/c) , (2/b) are in
$$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}\:\mathrm{then}\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}}\:\mathrm{are}\:\mathrm{in} \\ $$
Answered by Rasheed Soomro last updated on 04/Sep/16
a,b,c are in AP⇒b=((a+c)/2) (a/(bc)) , (1/c) , (2/b) (a/((((a+c)/2))c)) , (1/c) , (2/((a+c)/2)) ((2a)/(c(a+c))) , (1/c) , (4/(a+c)) (1/c)−((2a)/(c(a+c)))=^(?) (4/(a+c))−(1/c) ((a+c−2a)/(c(a+c)))=^(?) ((4c−a−c)/(c(a+c))) ((c−a)/(c(a+c)))≠((3c−a)/(c(a+c))) (a/(bc)) , (1/c) , (2/b) are not in AP (1/c)/((2a)/(c(a+c)))=^(?) (4/(a+c))/(1/c) (1/c)×((c(a+c))/(2a))=^(?) ((4c)/(a+c)) ((a+c)/(2a))≠((4c)/(a+c)) (a/(bc)) , (1/c) , (2/b) are not in GP ((2a)/(c(a+c))) , (1/c) , (4/(a+c)) 1/c is HM if above are in GP. So, 1/c =^(?) ((2(((2a)/(c(a+c))))((4/(a+c))))/(((2a)/(c(a+c)))+(4/(a+c)))) =^(?) (((16a)/(c(a+c)^2 ))/((2a+4c)/(c(a+c))))=(((8a)/(a+c))/(a+2c))=((8a)/((a+c)(a+2c))) ≠((8a)/((a+c)(a+2c))) Hence (a/(bc)) , (1/c) , (2/b) are not in HP.
$${a},{b},{c}\:\:{are}\:{in}\:{AP}\Rightarrow{b}=\frac{{a}+{c}}{\mathrm{2}} \\ $$$$\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}} \\ $$$$\:\frac{{a}}{\left(\frac{{a}+{c}}{\mathrm{2}}\right){c}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{\frac{{a}+{c}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\:\:,\:\frac{\mathrm{1}}{{c}}\:\:,\:\:\frac{\mathrm{4}}{{a}+{c}} \\ $$$$\frac{\mathrm{1}}{{c}}−\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\overset{?} {=}\frac{\mathrm{4}}{{a}+{c}}−\frac{\mathrm{1}}{{c}} \\ $$$$\frac{{a}+{c}−\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\overset{?} {=}\frac{\mathrm{4}{c}−{a}−{c}}{{c}\left({a}+{c}\right)} \\ $$$$\frac{{c}−{a}}{{c}\left({a}+{c}\right)}\neq\frac{\mathrm{3}{c}−{a}}{{c}\left({a}+{c}\right)} \\ $$$$\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}}\:{are}\:{not}\:{in}\:{AP} \\ $$$$\frac{\mathrm{1}}{{c}}/\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\overset{?} {=}\frac{\mathrm{4}}{{a}+{c}}/\frac{\mathrm{1}}{{c}} \\ $$$$\frac{\mathrm{1}}{{c}}×\frac{{c}\left({a}+{c}\right)}{\mathrm{2}{a}}\overset{?} {=}\frac{\mathrm{4}{c}}{{a}+{c}} \\ $$$$\frac{{a}+{c}}{\mathrm{2}{a}}\neq\frac{\mathrm{4}{c}}{{a}+{c}} \\ $$$$\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}}\:{are}\:{not}\:{in}\:{GP} \\ $$$$\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\:\:,\:\frac{\mathrm{1}}{{c}}\:\:,\:\:\frac{\mathrm{4}}{{a}+{c}} \\ $$$$\mathrm{1}/{c}\:\:{is}\:{HM}\:{if}\:{above}\:{are}\:{in}\:{GP}. \\ $$$${So},\:\:\:\:\:\mathrm{1}/{c}\:\overset{?} {=}\frac{\mathrm{2}\left(\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\right)\left(\frac{\mathrm{4}}{{a}+{c}}\right)}{\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}+\frac{\mathrm{4}}{{a}+{c}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{?} {=}\:\frac{\frac{\mathrm{16}{a}}{{c}\left({a}+{c}\right)^{\mathrm{2}} }}{\frac{\mathrm{2}{a}+\mathrm{4}{c}}{{c}\left({a}+{c}\right)}}=\frac{\frac{\mathrm{8}{a}}{{a}+{c}}}{{a}+\mathrm{2}{c}}=\frac{\mathrm{8}{a}}{\left({a}+{c}\right)\left({a}+\mathrm{2}{c}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\neq\frac{\mathrm{8}{a}}{\left({a}+{c}\right)\left({a}+\mathrm{2}{c}\right)} \\ $$$$\:{Hence}\:\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}}\:{are}\:{not}\:{in}\:{HP}. \\ $$$$ \\ $$

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