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Question Number 82628 by jagoll last updated on 23/Feb/20

$$Ifa,b,careinHarmonicprogression$$$$findthevalueof\frac{a+b}{b-a}+\frac{b+c}{b-c}.?$$

Commented by $@ty@m123 last updated on 23/Feb/20

$$\frac{1}{a},\frac{1}{b},\frac{1}{c}areinAP$$$$\Rightarrow \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}=\mathit{d},say$$$$\frac{a+b}{b-a}+\frac{b+c}{b-c}$$$$=\frac{a+b}{ab(\frac{1}{a}-\frac{1}{b})}+\frac{b+c}{bc(\frac{1}{c}-\frac{1}{b})}$$$$=-\frac{a+b}{ab\mathit{d}}+\frac{b+c}{bc\mathit{d}}$$$$=\frac{-c(a+b)+a(b+c)}{abc\mathit{d}}$$$$=\frac{b(a-c)}{abc\mathit{d}}$$$$=\frac{(a-c)}{ac\mathit{d}}$$$$=\frac{1}{\mathit{d}}(\frac{1}{c}-\frac{1}{a})$$$$=\frac{1}{\mathit{d}}\times (\mathit{d}+\frac{1}{b}+\mathit{d}-\frac{1}{b})$$$$=\frac{2\mathit{d}}{\mathit{d}}$$$$=2$$

Answered by TANMAY PANACEA last updated on 23/Feb/20

$$\frac{1}{a},\frac{1}{b},\frac{1}{c}areinA.P$$$$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\to 2=\frac{b}{a}+\frac{b}{c}$$$$\frac{a+b}{b-a}+\frac{b+c}{b-c}$$$$\frac{1+\frac{b}{a}}{\frac{b}{a}-1}+\frac{\frac{b}{c}+1}{\frac{b}{c}-1}$$$$\frac{1+\frac{b}{a}}{\frac{b}{a}-1}+\frac{2-\frac{b}{a}+1}{2-\frac{b}{a}-1}$$$$\frac{\frac{b}{a}+1}{\frac{b}{a}-1}+\frac{3-\frac{b}{a}}{1-\frac{b}{a}}$$$$\frac{\frac{b}{a}+1-3+\frac{b}{a}}{\frac{b}{a}-1}\to \frac{2(\frac{b}{a}-1)}{\frac{b}{a}-1}=2\mathit{a}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}$$

Commented by jagoll last updated on 23/Feb/20

$$thankyoumister$$

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