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28. If (b - c) ^ 2 , (c - a) ^ 2 (a - b) ^ 2 are in A.P. then (b - c) , (c - a) , (a - b) are in (a) A.P. (b)G .P (c) H.P. (d) None?
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28. If (b - c) ^ 2 , (c - a) ^ 2 (a - b) ^ 2 are in A.P. then (b - c) ...
Solution:

Given, (b - c) ^ 2 , (c - a) ^ 2 (a - b) ^ 2 are in A.P.

Let, (c - a) ^ 2 = (b - c) ^ 2 + d^2 ...(1) [as given in A.P.]

To prove: (b - c) , (c - a) ,(a - b) are in A.P.

Proof:

From equation (1), we get

(c - a) ^ 2 - (b - c) ^ 2 = d^2

On simplifying, we get

(c ^ 2 + a ^ 2 - 2ac) - (b ^ 2 + c ^ 2 - 2bc) = d^2

2c ^ 2 - 2ac - 2b ^ 2 + 2bc = d^2

c ^ 2 - ac - b ^ 2 + bc = d^2 / 2 ...(2)

Now, (b - c) ^ 2 , (c - a) ^ 2 (a - b) ^ 2 are in A.P.

Thus, (c - a) ^ 2 - (b - c) ^ 2 = (a - b) ^ 2 - (c - a) ^ 2

On simplifying, we get

2ac - 2bc = 2ab - 2ac

ac - bc = ab - ac

c(a + c) = b(a + c)

c = b [∵ a ≠ c]

Therefore, (b - c) = 0

Hence, (b - c) , (c - a) ,(a - b) are in A.P.

Therefore, the correct option is (a) A.P.
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28. If (b - c) ^ 2 , (c - a) ^ 2 (a - b) ^ 2 are in A.P. then (b - c) , (c - a) , (a - b) are in (a) A.P. (b)G .P (c) H.P. (d) None?
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28. If (b - c) ^ 2 , (c - a) ^ 2 (a - b) ^ 2 are in A.P. then (b - c) , (c - a) , (a - b) are in (a) A.P. (b)G .P (c) H.P. (d) None? for CA Foundation 2025 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about 28. If (b - c) ^ 2 , (c - a) ^ 2 (a - b) ^ 2 are in A.P. then (b - c) , (c - a) , (a - b) are in (a) A.P. (b)G .P (c) H.P. (d) None? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 28. If (b - c) ^ 2 , (c - a) ^ 2 (a - b) ^ 2 are in A.P. then (b - c) , (c - a) , (a - b) are in (a) A.P. (b)G .P (c) H.P. (d) None?.
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