text: # If (a – b), (b – c), (c – a) are in GP then prove that
  
  Question:
  
  If $(a-b),(b-c),(c-a)$ are in GP then prove that $(a+b+c)^{2}=3(a b+b c+$ ca).
  
  Solution:
  
  To prove: $(a+b+c)^{2}=3(a b+b c+c a)$.
  
  Given: $(a-b),(b-c),(c-a)$ are in GP
  
  Formula used: When $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}, \mathrm{b}^{2}=\mathrm{ac}$
  
  As, $(a-b),(b-c),(c-a)$ are in GP
  
  $\Rightarrow(b-c)^{2}=(a-b)(c-a)$
  
  $\Rightarrow b^{2}-2 c b+c^{2}=a c-a^{2}-b c+a b$
  
  $\Rightarrow a^{2}+b^{2}+c^{2}-b c-a c-a b=0$
  
  Adding 3(ab + bc + ac) both side
  
  $\Rightarrow a^{2}+b^{2}+c^{2}-b c-a c-a b+3(a b+b c+a c)=3(a b+b c+a c)$
  
  $\Rightarrow a^{2}+b^{2}+c^{2}+2 b c+2 a c+2 a b=3(a b+b c+a c)$
  
  $\Rightarrow(a+b+c)^{2}=3(a b+b c+a c)$
  
  Hence Proved
explanation: The document contains a math question and its corresponding solution. The question asks to prove that if \((a - b), (b - c), (c - a)\) are in geometric progression (GP), then \((a + b + c)^2 = 3(ab + bc + ca)\). The solution is provided and follows a step-by-step approach to prove the given statement.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # If (a – b), (b – c), (c – a) are in GP then prove that
  
  Question:
  
  If $(a-b),(b-c),(c-a)$ are in GP then prove that $(a+b+c)^{2}=3(a b+b c+$ ca).
  
  Solution:
  
  To prove: $(a+b+c)^{2}=3(a b+b c+c a)$.
  
  Given: $(a-b),(b-c),(c-a)$ are in GP
  
  Formula used: When $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}, \mathrm{b}^{2}=\mathrm{ac}$
  
  As, $(a-b),(b-c),(c-a)$ are in GP
  
  $\Rightarrow(b-c)^{2}=(a-b)(c-a)$
  
  $\Rightarrow b^{2}-2 c b+c^{2}=a c-a^{2}-b c+a b$
  
  $\Rightarrow a^{2}+b^{2}+c^{2}-b c-a c-a b=0$
  
  Adding 3(ab + bc + ac) both side
  
  $\Rightarrow a^{2}+b^{2}+c^{2}-b c-a c-a b+3(a b+b c+a c)=3(a b+b c+a c)$
  
  $\Rightarrow a^{2}+b^{2}+c^{2}+2 b c+2 a c+2 a b=3(a b+b c+a c)$
  
  $\Rightarrow(a+b+c)^{2}=3(a b+b c+a c)$
  
  Hence Proved
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and its corresponding solution. The question asks to prove that if \((a - b), (b - c), (c - a)\) are in geometric progression (GP), then \((a + b + c)^2 = 3(ab + bc + ca)\). The solution is provided and follows a step-by-step approach to prove the given statement.
  
  Label: question-and-answer