Question

show that b^2 is greater than, equal to, less than ac, according as a,b,c are in A.P., G.P or H.P

Answer 1

Answer:

Step-by-step explanation:

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)

Now,

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.

Answer 2

Step-by-step explanation:

only Hint for the given question..