Question

If a, b, c are in A.P., prove that the following are also in A.P.
$a^2(b+c),b^2(c+a),c^2(a+b)$ provided $\sum ab\ne 0$.

Answer 1

Given that $a,b,andc$ are in AP

$\therefore b-a=c-bor2b=a+c$ ......(i)

Now, let $a^2(b+c),b^2(c+a),and{c}^2(a+b)be{T}_1,T_2,T_{3}respectively$

$\therefore T_2-T_1=b^2(c+a)-a^2(b+c)$
$\Rightarrow c(b^2-a^2)+ab(b-a)=(b-a)(\sum ab)$
Similarly, $T_3-T_2=c^2(a+b)-b^2(c+a)=(c-b)(\sum ab)$
From (i), we know that $b-a=c-b$

$\therefore T_2-T_1=T_3-T_2$

$Hence,T_1,T_2,and{T}_{3}areinAP$