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when terms are in AP we know that 2B=A+C
therefore, 2[1/c+a)] = 1/b+c + 1/a+b
2/c+a = a+b+b+c/ (b+c)(a+b)
2/c+a = a+2b+c/ab+ac+b^2+bc
2ab+2ca+2b^2+2bc=2ca+2bc+c^2+a^2+2ab
2b^2=a^2+c^2
this is in the form of 2B=A+C
therefore, a^2,b^2,c^2 are in AP
therefore, 2[1/c+a)] = 1/b+c + 1/a+b
2/c+a = a+b+b+c/ (b+c)(a+b)
2/c+a = a+2b+c/ab+ac+b^2+bc
2ab+2ca+2b^2+2bc=2ca+2bc+c^2+a^2+2ab
2b^2=a^2+c^2
this is in the form of 2B=A+C
therefore, a^2,b^2,c^2 are in AP
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