If b+c a , c+a b , a+b b are A.P., prove that: 1 a , 1 b , 1 c are in A.P.

If 1 a ,1 b ,1 c are in A.P 1 b -1 a =1 c -1 b L.H.S=1 b -1 a = a-b ab = a(a-b) abc ...(i) R.H.S=1 c -1 b = a(b-c) abc ...(ii) Since, b+c a , c+a b , a+b c are in A.P b+c a - c+a b = c+a b - a+b c b^2+cd-ac-a^2 ab = c^2+ac-ab-b^2 bc (b-a)(a+b+c) ab = (c-d)(a+b+c) bc or a(a-c) abc = c(a-b) abc .....(3) From (1), (2) and (3) LHS=RHS Hence, 1 a , 1 b , 1 c are in A.P

If b+c a , c+a b , a+b b are A.P., prove that: 1 a , 1 b , 1 c ar…

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Question

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

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Answer

If $\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}$ are in A.P
$\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$\text{L.H.S}=\frac{1}{\text{b}}-\frac{1}{\text{a}}$
$=\frac{\text{a}-\text{b}}{\text{ab}}$
$=\frac{\text{a}(\text{a}-\text{b})}{\text{abc}}\ ...(\text{i})$
$\text{R.H.S}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$=\frac{\text{a}(\text{b}-\text{c})}{\text{abc}}\ ...(\text{ii})$
Since, $\frac{\text{b}+\text{c}}{\text{a}},\frac{\text{c}+\text{a}}{\text{b}},\frac{\text{a}+\text{b}}{\text{c}}$ are in A.P
$\frac{\text{b}+\text{c}}{\text{a}}-\frac{\text{c}+\text{a}}{\text{b}}=\frac{\text{c}+\text{a}}{\text{b}}-\frac{\text{a}+\text{b}}{\text{c}}$
$\frac{\text{b}^2+\text{cd}-\text{ac}-\text{a}^2}{\text{ab}}=\frac{\text{c}^2+\text{ac}-\text{ab}-\text{b}^2}{\text{bc}}$
$\Rightarrow\frac{(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})}{\text{ab}}=\frac{(\text{c}-\text{d})(\text{a}+\text{b}+\text{c})}{\text{bc}}$
or $\frac{\text{a}(\text{a}-\text{c})}{\text{abc}}=\frac{\text{c(a}-\text{b})}{\text{abc}}\ .....(3)$
From (1), (2) and (3)
$\text{LHS=RHS}$
Hence, $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P