If b - c -2- c - a - h - a - b - c are in A-P- prove that -pi 1-a- 1-h- 1-c are in A-P-ii bc- ca- ab are in A-P-

(i) 1/a, 1/b, 1/c are in A.P. Since b+c/a, c+a/b, a+b/c are in A.P., we have: c+a/b b+c/a=a+b/c c+a/b ⇒ac+a^2 b^2 bc/ab = ab+b^2 c^2 ac/bc ⇒(a+b)(a b)+c(a b)/ ...

Arithmetic Progression

If b + c /2, ...

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) bc, ca, ab are in A.P.

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Solution

(i) $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
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) bc, ca, ab are in A.P.
$\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \Rightarrow \frac{(a - b)}{a b} = \frac{(b - c)}{b c} \Rightarrow \frac{(a - b)}{a} = \frac{(b - c)}{c} \Rightarrow (a - b) c = a (b - c) \Rightarrow a c - b c = a b - a c$
Hence, bc, ca, ab are in A.P.

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Similar questions

Q. If

\frac{1}{a}, \frac{1}{b}, \frac{1}{c}

are in A.P., prove that
(i) bc,ca,ab are in A.P.
(ii) a(b+c), b(c+a), c(a+b) are in A.P

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\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}

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\frac{1}{a}, \frac{1}{b}, \frac{1}{c}

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\frac{1}{a}, \frac{1}{b}, \frac{1}{c}

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(i)

\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}

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If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., prove that:
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Q. If 1/a, 1/b, 1/c, are in a.p prove that BC, ca, an are in a.p
hints: prove that ab- ca =ca- BC
Using 1/c - 1/b = 1/b - 1/a.

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If b+ca,c+ab,a+bc are in A.P., prove that : (i) 1a,1b,1c are in A.P. (ii) bc, ca, ab are in A.P.

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) bc, ca, ab are in A.P.