If b - c - a - c - a - b - a - b - c are in A-P- prove thati 1-a- 1-b- 1-c are in A-P-ii bc- ca- ab are in A-P-

B+c/a,c+a/b,a+b/c are in A.P. b+c/a+1,c+a/b+1,a+b/c+1 are in A.P. b+c+a/a,c+a+b/b,a+b+c/c are in A.P. Dividing the terms by a+b+c We get, 1/a,1/b,1/c are in A. ...

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Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

If b + c / a ...

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

$\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in A.P.
$\frac{b + c}{a} + 1, \frac{c + a}{b} + 1, \frac{a + b}{c} + 1$ are in A.P.
$\frac{b + c + a}{a}, \frac{c + a + b}{b}, \frac{a + b + c}{c}$ are in A.P.
Dividing the terms by a+b+c
We get,
$\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
$\frac{a b c}{a} . \frac{a b c}{b}, \frac{a b c}{c}$ are in A.P.
$\Rightarrow$ bc, ac, 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

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.

Q. If

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

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

are in A.P.
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(ii) $a (b + c), b (c + a), c (a + b)$ are in A.P.

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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.

b+ca,c+ab,a+bc are in A.P. b+ca+1,c+ab+1,a+bc+1 are in A.P. b+c+aa,c+a+bb,a+b+cc are in A.P. Dividing the terms by a+b+c We get, 1a,1b,1c are in A.P. abca.abcb,abcc are in A.P. ⇒ bc, ac, 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.