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Question

If 1a,1b,1c are in A.P., prove that
(i) bc,ca,ab are in A.P.
(ii) a(b+c), b(c+a), c(a+b) are in A.P

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Solution

Given 1a,1b,1careinAP

(i)

1b1a=1c1b


(ab)ab=(bc)bc
(ab)a=(bc)c

acbc=abac

ac+ac=ab+bc

2ac=ab+bc

2ca=ab+bc

Thusbc,caandabareinAP

(ii)

We know that if, b(c + a) – a(b + c) = c(a + b) – b(c + a)

Consider LHS: b(c + a) – a(b + c)
Upon simplification we get,
b(c + a) – a(b + c) = bc + ba – ab – ac
= c (b - a)

Now, c(a + b) – b(c + a) = ca + cb – bc – ba
= a (c-b)

We know, 1a,1b,1careinAP

So, 1b1a=1c1b
Or c(b - a) = a(c - b)

Hence, given terms are in AP.


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