Question

If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., prove that that:

• A. $\frac{b+c}{a},\frac{+a}{b},\frac{a+b}{c},$ in A.P.
• B. $a(b+c),b(c+a),c(a+b)$ are in A.P.
• C. Both of these
• D. None of these

Answer 1

Answer: (B)

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

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

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

$\to 2ac=(a+c)b$

$2ac=ab+bc$

Add $ab+bc$ both sides

$\to 2ac+ab+bc=ab+ab+bc+bc$

$\to (ac+bc)+(ac+ab)=2(ab+bc)$

$\to c(a+b)+a(b+c)=2b(a+c)$

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