Question

40 points question please answer this question​

Answer 1

Note : x, y, z are in AP iff : x + z = 2y .......... (1) 

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∵ 1/a, 1/b, 1/c are in AP 

∴ (1/a) + (1/c) = 2(1/b) 

∴ (c+a) / (ca) = 2/b 

∴ b(c+a) = 2(ca) 

∴ (bc+ab) = 2(ca) ................... (2) 

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Then, we have : 

... a(b+c) + c(a+b) 

= ab + ca + ca + bc 

= (bc+ab) + 2(ca) 

= (bc+ab) + (bc+ab) ............ from (2) 

= 2(bc+ab) 

= 2 · b(c+a) 

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∴ a(b+c) + c(a+b) = 2 · b(c+a) 

∴ from (1), 

... a(b+c), b(c+a), c(a+b) are in AP. .................... Q.E.D. 

please mark this as brainlist

Answer 2

Step-by-step explanation:

Answer is in the attachment........

I hope that it will help you....