Question

Three unequal positive numbers a, b, c are in gp then prove that a + c >2b

Answer 1

Answer:

Since a, b, c are in GP

Therefore,

$b^2=ac$

$\implies b=\sqrt{ac}$  (Since a, b, c are positive, therefore negative value of b has been ignored)

We know that AM>GM

AM of a, and c > GM of a and c

Thus,

$\frac{a+c}{2}>\sqrt{ac}$

$\implies \frac{a+c}{2}>b$

$\implies a+c>2b$          (Proved)

Hope this is helpful.

Answer 2

Answer:

a + c > 2b

Step-by-step explanation:

Three unequal positive numbers a, b, c are in gp

let say three numbers are

a  , ar  , ar²

a = a

b = ar

c = ar²    

to be proved that

a + c > 2b

a + c > 2b

iff a + ar² > 2ar

iff a + ar² - 2ar > 0

iff a(1 + r² - 2r) > 0

iff a(1 - r)² > 0

a is a +ve number & square is always +ve

=>  a(1 - r)² > 0   if r≠1 and r can not be = 1 as a , b , c are unequal number

Hence Proved

a + c > 2b