Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

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.

$${a},{b},{c}\:\:{are}\:{in}\:{AP}\Rightarrow{b}=\frac{{a}+{c}}{\mathrm{2}} \\ $$$$\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}} \\ $$$$\:\frac{{a}}{\left(\frac{{a}+{c}}{\mathrm{2}}\right){c}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{\frac{{a}+{c}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\:\:,\:\frac{\mathrm{1}}{{c}}\:\:,\:\:\frac{\mathrm{4}}{{a}+{c}} \\ $$$$\frac{\mathrm{1}}{{c}}−\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\overset{?} {=}\frac{\mathrm{4}}{{a}+{c}}−\frac{\mathrm{1}}{{c}} \\ $$$$\frac{{a}+{c}−\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\overset{?} {=}\frac{\mathrm{4}{c}−{a}−{c}}{{c}\left({a}+{c}\right)} \\ $$$$\frac{{c}−{a}}{{c}\left({a}+{c}\right)}\neq\frac{\mathrm{3}{c}−{a}}{{c}\left({a}+{c}\right)} \\ $$$$\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}}\:{are}\:{not}\:{in}\:{AP} \\ $$$$\frac{\mathrm{1}}{{c}}/\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\overset{?} {=}\frac{\mathrm{4}}{{a}+{c}}/\frac{\mathrm{1}}{{c}} \\ $$$$\frac{\mathrm{1}}{{c}}×\frac{{c}\left({a}+{c}\right)}{\mathrm{2}{a}}\overset{?} {=}\frac{\mathrm{4}{c}}{{a}+{c}} \\ $$$$\frac{{a}+{c}}{\mathrm{2}{a}}\neq\frac{\mathrm{4}{c}}{{a}+{c}} \\ $$$$\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}}\:{are}\:{not}\:{in}\:{GP} \\ $$$$\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\:\:,\:\frac{\mathrm{1}}{{c}}\:\:,\:\:\frac{\mathrm{4}}{{a}+{c}} \\ $$$$\mathrm{1}/{c}\:\:{is}\:{HM}\:{if}\:{above}\:{are}\:{in}\:{GP}. \\ $$$${So},\:\:\:\:\:\mathrm{1}/{c}\:\overset{?} {=}\frac{\mathrm{2}\left(\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}\right)\left(\frac{\mathrm{4}}{{a}+{c}}\right)}{\frac{\mathrm{2}{a}}{{c}\left({a}+{c}\right)}+\frac{\mathrm{4}}{{a}+{c}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{?} {=}\:\frac{\frac{\mathrm{16}{a}}{{c}\left({a}+{c}\right)^{\mathrm{2}} }}{\frac{\mathrm{2}{a}+\mathrm{4}{c}}{{c}\left({a}+{c}\right)}}=\frac{\frac{\mathrm{8}{a}}{{a}+{c}}}{{a}+\mathrm{2}{c}}=\frac{\mathrm{8}{a}}{\left({a}+{c}\right)\left({a}+\mathrm{2}{c}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\neq\frac{\mathrm{8}{a}}{\left({a}+{c}\right)\left({a}+\mathrm{2}{c}\right)} \\ $$$$\:{Hence}\:\:\frac{{a}}{{bc}}\:,\:\frac{\mathrm{1}}{{c}}\:,\:\frac{\mathrm{2}}{{b}}\:{are}\:{not}\:{in}\:{HP}. \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com