Now this is a proving question
And can only be solved by showing all steps.
NO DIRECT ANSWER!!!!
And if you dont know then pls leave.
I dont want to know that you dont know it.
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Given that a, b, c are in A. P.
⇒ 2b = a + c ……. (1)
And a2, b2, c2 are in H. P.
⇒
⇒ (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)
i.e. a, b, -c/2 are in G. P. Hence ProvedPlz mark as brainliest if helpful
⇒ 2b = a + c ……. (1)
And a2, b2, c2 are in H. P.
⇒
⇒ (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)
i.e. a, b, -c/2 are in G. P. Hence ProvedPlz mark as brainliest if helpful
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