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in Algebra by (30.0k points)
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If a, b, c are in A.P show that following are also in A.P.

(i) \(\frac{1}{bc}\), \(\frac{1}{ca}\), \(\frac{1}{ab}\)

(ii) b + c, c + a, a + b

2 Answers

+2 votes
by (26.9k points)
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Best answer

(i) Given a, b, c are in A.P

⇒ \(\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}\) are also in A.P.

[On dividing each term by a, b, c]

⇒ \(\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}\) are also in A.P.

(ii) b + c, c + a, a + b will be in A.P.

If (c + a) − (b + c) = (a + b) − (c + a)

i.e if a − b = b − c

i.e if 2b = a + c

i.e if a, b, c are in A.P

Thus, a, b, c are in A.P ⇒ b + c, c + a, a + b are in A.P.

+2 votes
by (10.9k points)
since a, b, c are in A.P

we can write (a+c)=2b

Now 1/bc+1/ab=(a+c)/abc=2b/abc=2/ac

Hence 1/ab, 1/bc and 1/ca are in AP

Now (b+c)+(a+b)=2b+(c+a)=(c+a)+(c+a)=2(c+a)

So b+c,c+a and a+b are in AP

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