Question

prove that if a,b,c are in AP,then a(b+c)/bc,b(c+a)/ca and c(a+b)/ab are also in AP

Answer 1

subtract the first term from the second term and second term from the third term their difference will be equal to each other and if the difference comes equal than your answer is done your this is in AP otherwise not

Answer 2

If a, b and c are in AP

Then,

b-a=c-b ---1

Now,

b(c+a) /ca-a(b+c) /bc

=>[b(c+a) -{a(b+c)} ] /abc^2

=>bc-ac/ab^2

=>c(b-c) /abc^2

=>b-a/abc (from 1)---2

Again,

c(a+b) /ab -b(c+a) /ca

=>c(a+b) -{b(c+a)} /a^2bc

=>ac-ab/a^bc

=>b-c/abc

=>b-a/abc (from 1)---3

So, from 2 and 3 we can prove that

a(b+c) /bc, b(c+a) /ca and c(a+b) /ab are also in A. P.

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