If a b c are in A.P then show that ab(a+b),bc(b+c),ca(c+a) are also in A.P
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Answer:
∵a,b,careinAP
hence,(b−a)=(c−b)⟶(1)
(i) ∵b
2
(c+a)−a
2
(b+c)
=(b−a)(ab+bc+ca)
andc
2
(a+b)−b
2
(c+a)
=(c−b)(ab+bc+ca)
=(b−a)(ab+bc+ca)(from(1))
hence,b
2
(c+a)−a
2
(b+c)=c
2
(a+b)−b
2
(c+a)
∴a
2
(b+c),b
2
(c+a),c
2
(a+b)arealsoinAP
(ii) ∵(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)(from(1))
hence(c+a−b)−(b+c−a)=(a+b−c)−(c+a−b)
∴(b+c−a),(c+a−b),(a+b−c)areinAP
(iii) ∵(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)(from(1))
hence,(ca−b
2
)−(bc−a
2
)=(ab−c
2
)−(ca−b
2
)
∴(bc−a
2
),(ca−b
2
),(ab−c
2
)areinAP
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