Question
If 1b−c, 1c−a, 1a−b be consecutive terms of an A.P., then (b−c)2,(c−a)2,(a−b)2 will be in ___.
Answer: Option B
:
B
If(b−c)2,(c−a)2,(a−b)2 are in A.P.
Then we have(c−a)2−(b−c)2 = (a−b)2−(c−a)2
⇒(b−a)(2c−a−b) = (c - b)(2a - b - c) ......(i)
Also if 1b−c,1c−a,1a−b are in A.P.
Then 1c−a−1b−c = 1a−b−1c−a
⇒b+a−2c(c−a)(b−c) = c+b−2a(a−b)(c−a)
⇒(a−b)(b+a−2c) = (b−c)(c+b−2a)
⇒(b−a)(2c−a−b) = (c−b)(2a−b−c)
Which is true by virtue of (i).
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:
B
If(b−c)2,(c−a)2,(a−b)2 are in A.P.
Then we have(c−a)2−(b−c)2 = (a−b)2−(c−a)2
⇒(b−a)(2c−a−b) = (c - b)(2a - b - c) ......(i)
Also if 1b−c,1c−a,1a−b are in A.P.
Then 1c−a−1b−c = 1a−b−1c−a
⇒b+a−2c(c−a)(b−c) = c+b−2a(a−b)(c−a)
⇒(a−b)(b+a−2c) = (b−c)(c+b−2a)
⇒(b−a)(2c−a−b) = (c−b)(2a−b−c)
Which is true by virtue of (i).
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