Question

if b+c , c+a , a+b are in AP , then show that a(1/b+1/c) , b(1/c+1/a) , c(1/a+1/b) are also in AP
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Answer 1

Answer:

As, these three values are in A.P.

∴b(1c+1a)−a(1b+1c)=c(1a+1b)−b(1c+1a)

⇒bc+ba−ab−ac=ca+cb−bc−ba

⇒1c(b−a)+(b2−a2ab)=1a(c−b)+(c2−b2bc)

⇒1c(b−a)+(b−a)(b+a)ab)=1a(c−b)+(c−b)(c+b)bc)

⇒(b−a)(1c+b+aab)=(c−b)(1a+c+bcb)

⇒(b−a)(1c+1a+1b)=(c−b)(1a+1b+1c)

⇒b−a=c−b

So, a,bandc are in AP.

Step-by-step explanation:

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