Show that b^2 is greater than equal to or less than ac, according as a, b, c are in A. P., G. P. Or HP
arithmetic mean is 25 and geometric mean is 20.

Answer/Step-by-step explanation  (ac > b² or b² < ac. )

A/c to question, we have to show:-

b² >ac in A.P (1)

b² = ac in G.P (2)

b² < ac in H.P. (3)

b = a+c/2 (A.P)

b = √ac ( G.P)

b = 2ac/a+c (H.P)

In A.P :

b² > ac = b² - ac

= (a+c/2)² - ac

= (a²+2ac+c²/4) - ac  = a² + 2ac + c² - 4ac / 4

= a² - 2ac + c² / 4  = ( a - c ) ² / 4 > 0  Hence, b²>ac

In G.P:-

b = √ac

Hence, b² = ac

In H.P :-  b² < ac = ac > b²  = ac - b²  = ac - ( 2ac / a+c)

= ac(a+c) - 2ac / a+c

= a²c + ac² - 2ac / a+c

= ac(2ac - 2) / a+c > 0

Hence, ac > b² or b² < ac.

i dont have an answer

step-by-step explanation: