Question

If $\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in $A.P.,$ then which of the following is in $A.P.?$

• A. $a,b,c$
• B. $a^2,b^2,c^2$
• C. $\frac{1}{a},\frac{1}{b},\frac{1}{c}$
• D. $bc,ac,ab$

Answer 1

Answer: (C), (D)

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$
$bc,ac,ab$

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

Adding $2$ to each term

$\frac{b+c-a}{a}+2,\frac{c+a-b}{b}+2,\frac{a+b-c}{c}+2$ are in A.P.

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

Dividing each term by $(a+b+c)$,

$\frac{a+b+c}{a(a+b+c)},\frac{a+b+c}{b(a+b+c)},\frac{a+b+c}{c(a+b+c)}$ are in A.P,

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

Multiplying each term by $abc$

$\frac{abc}{a},\frac{abc}{b},\frac{abc}{c}$ are in A.P.

$\therefore bc,ac,ab$ are in A.P.