If b-c-a-c-a-b-a-b-c are in AP- then prove that i 1-a-1-b-1-care in AP-ii bc-ca-abare in AP-

Step 1. Note the given data and adding 1 with each term of the A.P.Given that b+c/a,c+a/b,a+b/c are in A.P.Adding 1 with each term of the A.P.b+c/a+1,c+a/b+1,a+ ...

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Quantitative Aptitude

Quadratic Equations

If b+c/a,c+a/...

Question

If $\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in AP, then prove that
(i) $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP.
(ii) $b c, c a, a b$ are in AP.

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Solution

Step 1. Note the given data and adding $1$ with each term of the A.P.
Given that
$\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in A.P.
Adding $1$ with each term of the A.P.
$\frac{b + c}{a} + 1, \frac{c + a}{b} + 1, \frac{a + b}{c} + 1$ are in A.P.
$\Rightarrow \frac{b + c}{a} + \frac{a}{a}, \frac{c + a}{b} + \frac{b}{b}, \frac{a + b}{c} + \frac{c}{c}$ are in A.P.
$\Rightarrow \frac{b + c + a}{a}, \frac{c + a + b}{b}, \frac{a + b + c}{c}$ are in A.P.
Step 2. Dividing each term by $(a + b + c)$
$\frac{b + c + a}{a (a + b + c)}, \frac{c + a + b}{b (a + b + c)}, \frac{a + b + c}{c (a + b + c)}$
$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
$∴ \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. …………(i)
(i) Hence, $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
Step 3. Multiply $a b c$ with each term in A.P. $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$
$\frac{a b c}{a}, \frac{a b c}{b}, \frac{a b c}{c}$ are in A.P.
$\Rightarrow b c, a c, a b$ are in A.P.
(ii) Hence $b c, a c, a b$ are in A.P.

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If b+ca,c+ab,a+bc are in AP, then prove that (i) 1a,1b,1care in AP.(ii) bc,ca,abare in AP.

If $\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in AP, then prove that
(i) $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP.
(ii) $b c, c a, a b$ are in AP.