Question

If
$\frac{b + c}{a} \: and \: \frac{c + a}{b} and \: \frac{a + b}{c}$
Are in A.P.
Then prove that
$\frac{1}{a} \: and \: \frac{1}{b} \: and \: \frac{1}{c}$
Are in A.P.( Arithmetic progression)​

Answer 1

Step-by-step explanation:

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

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

$\Longrightarrow \frac{a(a + c) - b(b + c)}{ab} =\frac{b(a + b) - c(a + c)}{bc}$

$\Longrightarrow \frac{a^2 + ac - b^2 - bc}{ab} =\frac{ab+b^2-ac-c^2}{bc}$

$\Longrightarrow \frac{c(a-b) + (a^2-b^2)}{ab} =\frac{a(b-c) + (a^2 - b^2)}{bc}$

$\Longrightarrow \frac{c(a - b) + (a + b)(a - b)}{ab} = \frac{a(b - c) + (a + b)(a - b)}{bc}$

$\Longrightarrow \frac{(a-b)[a+b+c]}{ab} = \frac{(b-c)[a+b+c]}{bc}$

$\Longrightarrow \frac{a-b}{ab} = \frac{b-c}{bc}$

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

$\Longrightarrow \frac{2}{b} = \frac{1}{a}+\frac{1}{c}$

$\Longrightarrow \boxed{ {\frac{1}{a}, \ \frac{1}{b}, \ \frac{1}{c}\ are \ in \ AP}}$

Hope it helps!

Answer 2

iven = 1/a+b, 1/b+c, 1/c+a

Proof = 1/b+c-1/a+b=1/c+a-1/b+c

a+b-(b+c)/(b+c)(a+b) = b+c-(c+a)/(b+c)(c+a)

a-c/a+b=b-a/c+a

a-c/a+b=b-a/c+a

(a-c)(a+c)=(b-a)(b+a)

a²-c²=b²-a²

c², a², b²