Question

If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., prove that:
(i) $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P.
(ii) a (b +c), b (c + a), c (a +b) are in A.P.

Answer 1

$$\begin{gathered} \mathrm{Given}:\frac{1}{a},\frac{1}{b},\frac{1}{c}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore \frac{2}{b}=\frac{1}{a}+\frac{1}{c} \\ \Rightarrow 2ac=ab+bc....\left(1\right) \\ \left(\mathrm{i}\right)\mathrm{To}\mathrm{prove}:\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ 2\left(\frac{a+c}{b}\right)=\frac{b+c}{a}+\frac{a+b}{c} \\ \Rightarrow 2ac(a+c)=bc(b+c)+ab(a+b) \\ \mathrm{LHS}:2ac(a+c) \\ =(ab+bc)(a+c)\left(\mathrm{From}\right(1\left)\right) \\ =a^{2}b+2abc+b{c}^2 \\ \mathrm{RHS}:bc(b+c)+ab(a+b) \\ =b^{2}c+b{c}^2+a^{2}b+a{b}^2 \\ =b^{2}c+a{b}^2+b{c}^2+a^{2}b \\ =b(bc+ab)+b{c}^2+a^{2}b \\ =2abc+b{c}^2+a^{2}b \\ =a^{2}b+2abc+b{c}^2\left(\mathrm{From}\right(1\left)\right) \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{To}\mathrm{prove}:\mathrm{a}(\mathrm{b}+\mathrm{c}),\mathrm{b}(\mathrm{c}+\mathrm{a}),\mathrm{c}(\mathrm{a}+\mathrm{b})\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \Rightarrow 2b(c+a)=a(b+c)+c(a+b) \\ \mathrm{LHS}:2b(c+a) \\ =2bc+2ba \\ \mathrm{RHS}:a(b+c)+c(a+b) \\ =ab+ac+ac+bc \\ =ab+2ac+bc \\ =ab+ab+bc+bc\left(\mathrm{From}\right(1\left)\right) \\ =2ab+2bc \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$