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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 |
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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. since a, b, c are in A.Pwe 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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