If $a, b, c$ are in A.P., prove that $b + c, c + a, a + b$ are also in A.P.
Step-by-Step Solution
Key Concept: Subtract $(a + b + c)$ from each term: $(b+c) - (a+b+c) = -a$, $(c+a) - (a+b+c) = -b$, $(a+b) - (a+b+c) = -c$. Since $a, b, c$ are in A.P., $-a, -b, -c$ are in A.P., so $b+c, c+a, a+b$ are in A.P.
Let $b+c, c+a, a+b$ be in A.P. Common difference check: $(c+a) - (b+c) = a - b$. [1.0 Mark]
$(a+b) - (c+a) = b - c$. Since $a, b, c$ are in A.P., $b - a = c - b \Rightarrow a - b = b - c$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding common difference $a - b$ and $b - c$: 1.0 Mark
Equating using A.P. definition: 1.0 Mark
Correct Answer: