Question

any one. 100 points question

Answer 1

Go through the following attachment

Answer 2

Here is your answer
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Given that a, b, c are in A. P.
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⇒ 2b = a + c. = ……. (1) =

And a2, b2, c2 are in H. P.

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⇒ 
$\frac{1}{b {}^{2} } - \frac{1}{d {}^{2} } - \frac{1}{c {}^{2} } - \frac{1}{b {}^{2} }$


⇒ (a – b)(a + b)/b2a2

 = (b – c) (b + c)/b2c2

⇒ ac2 + bc2 = a2b + a2c [∵ a – b = b – c]

⇒ ac (c – a) + b (c – a) (c + a)
= 0

⇒ (c – a) (ab + bc + ca)
= 0

⇒ either c – a = 0 or ab + bc + ca
= 0

⇒ either c = a or (a+ c) b + ca
= 0 and then form (i) 2b2 + ca
= 0

Either a = b = c or b2 
= a (-c/2)

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a, b, -c/2 are in G. P. Hence Proved