Question

If $a(\frac{1}{b}+\frac{1}{c}),b(\frac{1}{c}+\frac{1}{a}),c(\frac{1}{a}+\frac{1}{b})$ are in A.P., then $a,b,c$ are also

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. None of these

Answer 1

Answer: (A)

$A.P.$

If $a,b,c$ are in A.P then

$2b=a+c$___________ (1)

$\therefore 2b(\frac{1}{c}+\frac{1}{a})=a(\frac{1}{b}+\frac{1}{c})+(\frac{1}{a}+\frac{1}{b})c$

$\frac{2b(a+c)}{ac}=\frac{a(b+c)}{bc}+\frac{(b+a)}{ab}c$

Multiplying by $a,b,c$

$2b^2(a+c)=a^2(b+c)+c^2(a+b)$

If a,b,c are in A.P, then it should satisfy above equation

$LHS=2b^2(a+c)$

$=2b^2\left(2b\right)=4b^3$ [ from $\left(i\right)$]

$RHS=a^2(b+c)+c^2(a+b)$

$=a^2(\frac{a+c}{2}+c)+c^2(a+\frac{a+c}{2})$

$=a^2\left(\frac{a+3c}{2}\right)+c^2\left(\frac{3a+c}{2}\right)$

$=\frac{a^3+3a^{2}c+3c^{2}a+c^3}{2}$

$=\frac{(a+c{)}^3}{2}$

$=\frac{(2b{)}^3}{2}$

$=4b^3$ Thus $a,b,c$ are in A.P

$=LHS$