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Question Number 82628 by jagoll last updated on 23/Feb/20
If a,b, c are in Harmonic progression find the value of ((a+b)/(b−a)) + ((b+c)/(b−c)) . ?
$${If}\:{a},{b},\:{c}\:{are}\:{in}\:{Harmonic}\:{progression} \\ $$$${find}\:{the}\:{value}\:{of}\:\frac{{a}+{b}}{{b}−{a}}\:+\:\frac{{b}+{c}}{{b}−{c}}\:.\:? \\ $$
Commented by $@ty@m123 last updated on 23/Feb/20
(1/a), (1/b),(1/c) are in AP ⇒(1/b)−(1/a)=(1/c)−(1/b)=d, say ((a+b)/(b−a)) + ((b+c)/(b−c)) =((a+b)/(ab((1/a)−(1/b))))+((b+c)/(bc((1/c)−(1/b)))) =−((a+b)/(abd))+((b+c)/(bcd)) =((−c(a+b)+a(b+c))/(abcd)) =((b(a−c))/(abcd)) =(((a−c))/(acd)) =(1/d)((1/c)−(1/a)) =(1/d)×(d+(1/b)+d−(1/b)) =((2d)/d) =2
$$\frac{\mathrm{1}}{{a}},\:\frac{\mathrm{1}}{{b}},\frac{\mathrm{1}}{{c}}\:{are}\:{in}\:{AP} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{{b}}=\boldsymbol{{d}},\:{say} \\ $$$$\frac{{a}+{b}}{{b}−{a}}\:+\:\frac{{b}+{c}}{{b}−{c}}\: \\ $$$$=\frac{{a}+{b}}{{ab}\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)}+\frac{{b}+{c}}{{bc}\left(\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{{b}}\right)} \\ $$$$=−\frac{{a}+{b}}{{ab}\boldsymbol{{d}}}+\frac{{b}+{c}}{{bc}\boldsymbol{{d}}} \\ $$$$=\frac{−{c}\left({a}+{b}\right)+{a}\left({b}+{c}\right)}{{abc}\boldsymbol{{d}}} \\ $$$$=\frac{{b}\left({a}−{c}\right)}{{abc}\boldsymbol{{d}}} \\ $$$$=\frac{\left({a}−{c}\right)}{{ac}\boldsymbol{{d}}} \\ $$$$=\frac{\mathrm{1}}{\boldsymbol{{d}}}\left(\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{{a}}\right) \\ $$$$=\frac{\mathrm{1}}{\boldsymbol{{d}}}×\left(\boldsymbol{{d}}+\frac{\mathrm{1}}{{b}}+\boldsymbol{{d}}−\frac{\mathrm{1}}{{b}}\right) \\ $$$$=\frac{\mathrm{2}\boldsymbol{{d}}}{\boldsymbol{{d}}} \\ $$$$=\mathrm{2} \\ $$
Answered by TANMAY PANACEA last updated on 23/Feb/20
(1/a),(1/b),(1/c) are in A.P (2/b)=(1/a)+(1/c) →2=(b/a)+(b/c) ((a+b)/(b−a))+((b+c)/(b−c)) ((1+(b/a))/((b/a)−1))+(((b/c)+1)/((b/c)−1)) ((1+(b/a))/((b/a)−1))+((2−(b/a)+1)/(2−(b/a)−1)) (((b/a)+1)/((b/a)−1))+((3−(b/a))/(1−(b/a))) (((b/a)+1−3+(b/a))/((b/a)−1)) →((2((b/a)−1))/((b/a)−1))=2 answer
$$\frac{\mathrm{1}}{{a}},\frac{\mathrm{1}}{{b}},\frac{\mathrm{1}}{{c}}\:\:\:{are}\:{in}\:{A}.{P} \\ $$$$\frac{\mathrm{2}}{{b}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}\:\:\:\rightarrow\mathrm{2}=\frac{{b}}{{a}}+\frac{{b}}{{c}} \\ $$$$\frac{{a}+{b}}{{b}−{a}}+\frac{{b}+{c}}{{b}−{c}} \\ $$$$\frac{\mathrm{1}+\frac{{b}}{{a}}}{\frac{{b}}{{a}}−\mathrm{1}}+\frac{\frac{{b}}{{c}}+\mathrm{1}}{\frac{{b}}{{c}}−\mathrm{1}} \\ $$$$\frac{\mathrm{1}+\frac{{b}}{{a}}}{\frac{{b}}{{a}}−\mathrm{1}}+\frac{\mathrm{2}−\frac{{b}}{{a}}+\mathrm{1}}{\mathrm{2}−\frac{{b}}{{a}}−\mathrm{1}} \\ $$$$\frac{\frac{{b}}{{a}}+\mathrm{1}}{\frac{{b}}{{a}}−\mathrm{1}}+\frac{\mathrm{3}−\frac{{b}}{{a}}}{\mathrm{1}−\frac{{b}}{{a}}} \\ $$$$\frac{\frac{{b}}{{a}}+\mathrm{1}−\mathrm{3}+\frac{{b}}{{a}}}{\frac{{b}}{{a}}−\mathrm{1}}\:\rightarrow\frac{\mathrm{2}\left(\frac{{b}}{{a}}−\mathrm{1}\right)}{\frac{{b}}{{a}}−\mathrm{1}}=\mathrm{2}\:\:\:\boldsymbol{{answer}} \\ $$
Commented by jagoll last updated on 23/Feb/20
thank you mister
$${thank}\:{you}\:{mister} \\ $$

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