Solve the following

Question:

If $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P., prove that:

(i) $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

(ii) bccaab are in A.P.

Solution:

(i) Since $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P., we have:

$\frac{c+a}{b}-\frac{b+c}{a}=\frac{a+b}{c}-\frac{c+a}{b}$

$\Rightarrow \frac{a c+a^{2}-b^{2}-b c}{a b}=\frac{a b+b^{2}-c^{2}-a c}{b c}$

$\Rightarrow \frac{(a+b)(a-b)+c(a-b)}{a b}=\frac{(b+c)(b-c)+a(b-c)}{b c}$

$\Rightarrow \frac{(a-b)(a+b+c)}{a b}=\frac{(b-c)(a+b+c)}{b c}$

$\Rightarrow \frac{(a-b)}{a b}=\frac{(b-c)}{b c}$

$\Rightarrow \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

Hence, $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

(ii) Since $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., we have:

$\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

$\Rightarrow \frac{(a-b)}{a b}=\frac{(b-c)}{b c}$

$\Rightarrow \frac{(a-b)}{a}=\frac{(b-c)}{c}$

$\Rightarrow(a-b) c=a(b-c)$

$\Rightarrow a c-b c=a b-a c$

Hence, bccaab are in A.P.

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