Question

if b-c, 2b-x and b-a are in H.P then a-(x/2), b-(x/2), c-(x/2) are in a)A.P b)G.P c)H.P d)none

Answer 1

Answer:

a-(x/2), b-(x/2), c-(x/2) are in a GP

Step-by-step explanation:

b-c, 2b-x and b-a are in H.P

=> 2/(2b - x)  = 1/(b-c)  + 1/(b-a)

=> 2/(2b - x) = (b - a + b - c) /(b-c)(b-a)

=> 2 (b-c)(b-a) = (2b - a - c) (2b - x)

=> 2b² - 2bc - 2ab + 2ac = 4b² - 2ab - 2bc  - 2bx  + ax + cx

=> 2ac  = 2b² - 2bx  + ax + cx

=> ac = b² - bx + ax/2  + cx/2

=> ac - ax/2 - cx/2 = b² - bx

Adding x²/4 both sides

=> ac - ax/2 - cx/2 + x²/4  = b² - bx + x²/4

=> a(c - x/2) - (x/2)(c - x/2) = b² -bx/2 - bx/2 + x²/4

=> a(c - x/2) - (x/2)(c - x/2) = b(b - x/2) - x/2(b - x/2)

=> (a -  x/2)(c-   x/2) =  (b - x/2)²

=>  a-(x/2), b-(x/2), c-(x/2) are in a GP