Question

if 1/b+c,1/c+a,1/a+B are in AP then prove 2b^2=a^2+c^2​

Answer 1

Answer:

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Step-by-step explanation:

since 1/b+c,1/c+a,1/a+b are in A.P.

so,

[1/(c+a)]-[1/(b+c)]=[1/(a+b)]-[1/(c+a)]

[(b+c-c-a)/(b+c)(c+a)]=[(c+a-a-b)/(c+a)(a+b)]

(b-a)/(b+c)=(c-b)/(a+b)

b^2-a^2=c^2-b^2

2b^2=c^2+a^2

Hence Proved.

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