You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Sum of binomial coefficients of even numbered terms

If A and G ar...

Question

If A and G are respectively the arithmetic and geometric means between two distinct positive numbers a and b then prove that A>G.

Open in App

Solution

We have
$A = \frac{a + b}{2} and G = \sqrt{a b} \Rightarrow A - G = \frac{a + b}{2} - \sqrt{a b} = \frac{a + b - 2 \sqrt{a b}}{2} = \frac{1}{2} (\sqrt{a} - \sqrt{b})^{2} > 0 \Rightarrow A > G . Hence, A M > G M .$

Suggest Corrections

Similar questions

if the arithmetic, geometric and harmonic means between two distinct positive real numbers be A, G and H respectively, then the relation between them is

Q. If

A

and

G

are the Arithmetic mean and Geometric mean respectively of two numbers, then prove that the numbers are

A \pm \sqrt{(A + G) (A - G)}

Q. Let

a

and

b

be two positive real numbers. Suppose

A_{1}, A_{2}

are two arithmetic means;

G_{1}, G_{2}

are two geometric means and

H_{1}, H_{2}

are two harmonic means between

a

and

b

, then-

Q. If

\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}

represents arithmetic mean, geometric mean, harmonic mean between two distinct positive numbers

a

and

b

for

n = α, β, γ

respectively, then the equation whose roots are

α, 2 β, γ

Q. If the arithmetic and geometric mean of two positive real numbers

a

and

b (a > b)

are

13

and

12

respectively, then the value of

a - b

Join BYJU'S Learning Program

If A and G are respectively the arithmetic and geometric means between two distinct positive numbers a and b then prove that A>G.

We have A=a+b2andG=√ab⇒A−G=a+b2−√ab=a+b−2√ab2=12(√a−√b)2>0⇒A>G.Hence,AM>GM.

We have
$A = \frac{a + b}{2} and G = \sqrt{a b} \Rightarrow A - G = \frac{a + b}{2} - \sqrt{a b} = \frac{a + b - 2 \sqrt{a b}}{2} = \frac{1}{2} (\sqrt{a} - \sqrt{b})^{2} > 0 \Rightarrow A > G . Hence, A M > G M .$