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 are in HP, then which of the following is/are correct?
1. a, b, c are in AP
2. (b + c)2, (c + a)2, (a + b)are in GP. Select the correct answer using the code given below.
  • a)
    1 only
  • b)
    2 only
  • c)
    Both 1 and 2
  • d)
    Neither 1 nor 2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
are in HP, then which of the following is/are correct?1. a, b, c are i...
Concept:
  • If a, b and c are three terms in GP then b2 = ab and vice-versa
  • If three terms a, b, c are in HP, then and vice-versa
  • When three quantities are in AP, the middle one is called as the arithmetic mean of the other two.
  • If a, b and c are three terms in AP then 
  • and vice-versa
  • If a, b and c are in A.P then 1/a, 1/b, 1/c are in H.P and vice-versa
Calculation:
(b + c), (c + a) and (a + b) are in A.P
⇒ 2(c + a) = b + c + a + b    
⇒ 2c + 2a = 2b + a + c     -----(i)
⇒ 2c + 2a - 2b - a - c = 0
⇒ 2b = c + a
So, a, b, c are in AP
Now, Let (b + c)2, (c + a)2, (a + b)2 are in GP
(c + a)2 = √[(b + c).(a + b)]2
⇒ c2 + a2 + 2ac = (b + c).(a + b)
⇒ c2 + a2 + 2ac = ab + b2 + ac + bc
⇒ c2 + a2 - b2 + ac - ab - bc = 0
From here we are unable to check the relation between a, b and c 
So, Our assumption was wrong 
So, (b + c)2, (c + a)2, (a + b)2 are not in GP.
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