If a,b,care in AP and a²,b²,c² are in HP, then which of the following could hold true?
Answers
Answered by
3
a^2,b^2,c^2 are in H.P
=> 1/b^2 - 1/d^2 = 1/c^2 - 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)
i.e. a, b, -c/2 are in G. P. Hence Proved
Similar questions