Question

If a, b, c are in A.P., then show that:
(i) $a^2(b+c),b^2(c+a),c^2(a+b)$ are also in A.P.
(ii) $b+c-a,c+a-b,a+b-c$ are in A.P.
(iii) $bc-a^2,ca-b^2,ab-c^2$ are in A.P.

Answer 1

Since a, b, c are in A.P., we have:
2b = a + c
(i) $a^2(b+c),b^2(c+a),c^2(a+b)$ are also in A.P.
We have to prove that following:
$2b^2(a+c)=a^2(b+c)+c^2(a+b)$
LHS: $2b^2\times 2b$ (Given)
$=4b^3$
RHS: $a^{2}b+a^{2}c+a{c}^2+c^{2}b$
$=ac(a+c)+b(a^2+c^2)=ac(a+c)+b\left[\right(a+c{)}^2-2ac]=ac(2b)+b[(2b{)}^2-2ac]=2abc+4b^3-2abc=4b^3$
LHS = RHS

(ii) $b+c-a,c+a-b,a+b-c$ are in A.P.
We have to prove that following:
$2(c+a-b)=(b+c-a)+(a+b-c)$
LHS: $2(c+a-b)$
$=2(2b-b)$ $(\because 2b=a+c)$
$=2b$
RHS: $(b+c-a)+(a+b-c)$
$=2b$
LHS = RHS

(iii) $bc-a^2,ca-b^2,ab-c^2$ are in A.P.
We have to prove that
$2(ca-b^2)=(bc-a^2+ab-c^2)$
RHS: $bc-a^2+ab-c^2$
$=c(b-c)+a(b-a)=c(\frac{a+c}{2}-c)+a(\frac{a+c}{2}-a)(\because 2b=a+c)$
$=c\left(\frac{a+c-2c}{2}\right)+a\left(\frac{a+c-2a}{2}\right)=\frac{c(a-c)}{2}+a\left(\frac{c-a}{2}\right)=\frac{ca}{2}-\frac{c^2}{2}+\frac{ac}{2}-\frac{a^2}{2}=ac-\frac{1}{2}(c^2+a^2)=ac-\frac{1}{2}(4b^2-2ac)(\because a^2+c^2+2ac=4b^2\Rightarrow a^2+c^2=4b^2-2ac)=ac-2b^2+ac=2ac-2b^2=2(ac-b^2)$
= LHS
Hence proved.