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

$\because a,b,careinAPhence,(b-a)=(c-b)⟶\left(1\right)$

(i) $\because b^2(c+a)-a^2(b+c)=(b-a)(ab+bc+ca)and{c}^2(a+b)-b^2(c+a)=(c-b)(ab+bc+ca)=(b-a)(ab+bc+ca)\left(from\right(1\left)\right)hence,b^2(c+a)-a^2(b+c)=c^2(a+b)-b^2(c+a)\therefore a^2(b+c),b^2(c+a),c^2(a+b)arealsoinAP$

(ii) $\because (c+a-b)-(b+c-a)=2(a-b)=-2(b-a)and(a+b-c)-(c+a-b)=2(b-c)=-2(c-b)=-2(b-a)\left(from\right(1\left)\right)hence(c+a-b)-(b+c-a)=(a+b-c)-(c+a-b)\therefore (b+c-a),(c+a-b),(a+b-c)areinAP$

(iii) $\because (ca-b^2)-(bc-a^2)=(a-b)(a+b+c)=-(b-a)(a+b+c)and(ab-c^2)-(ca-b^2)=(b-c)(a+b+c)=-(c-b)(a+b+c)=-(b-a)(a+b+c)\left(from\right(1\left)\right)hence,(ca-b^2)-(bc-a^2)=(ab-c^2)-(ca-b^2)\therefore (bc-a^2),(ca-b^2),(ab-c^2)areinAP$