Question

If $a,b,c$ are in A.P., then show that,
$a^2(b+c);b^2(c+a),c^2(a+b)$ are in $A$ $(ab+bc+ca\ne )0$

Answer 1

if $a^2(b+c),b^2(a+c),c^2(a+b)$ are in AP then
$2b^2(a+c)=a^2(b+c)+c^2(a+b)$ (we have to prove this)
RHS:
$=a^2(b+c)+c^2(a+b)$
$=a^{2}b+a^{2}c+c^{2}a+c^{2}b$
$=b(a^2+c^2)+ac(a+c)$
since a,b,c are in AP so $b=\frac{a+c}{2}$
RHS $=\frac{(a+c)(a^2+b^2)}{2}+ac(a+c)$
$=\frac{(a+c)(a^2+b^2+2ac)}{2}$
$=\frac{(a+c{)}^3}{2}$
$=2(a+c)b^2$ (after putting $a+c=2b$)