If $G_{1}$ and $G_{2}$ are two geometric means and $A$ is the arithmetic mean inserted between two positive numbers , then the value of $\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}$ is

4 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 A$
Now if two geometric means are inserted between $a^{3}$ and $b^{3}$ that means $b^{3}$ is the fourth term of that GP.
So $a^{3} r^{3} = b^{3}$
$r = b / a$
$G 1 = a^{3} r = a^{2} b$
$G 2 = a^{3} r^{2} = a b^{2}$
$(G 1)^{2} / G 2 + (G 2)^{2} / G 1 = [a^{4} b^{2} / a b^{2}] + [a^{2} b^{4} / a^{2} b] = a^{3} + b^{3} = 2 A$

Suggest Corrections

Similar questions

If one A.M., A and two geometric means $G_{1}$ and $G_{2}$ inserted between any two positive numbers, show that $\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}} = 2 A$ .

Q. If A is the arithmetic mean and

G_{1}, G_{2}

be two geometric means between any two numbers, then

\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}

is equal to :

Q. Let

3

geometric means

G_{1}, G_{2}, G_{3}

are inserted between two positive number

a

and

b

such that

\frac{G_{3} - G_{2}}{G_{2} - G_{1}} = 2

, then

\frac{b}{a}

equal to

Q. If

A_{1}, A_{2}

be two arithmetic means and

G_{1}, G_{2}

be two geometric means between two positive numbers a and b, then

\frac{A_{1} + A_{2}}{G_{1} G_{2}}

Q. If between any two quantities there be inserted two arithmetic means

A_{1}, A_{2}

; two geometric means

G_{1}, G_{2} : H_{1} H_{2} = A_{1} + A_{2} : H_{1} + H_{2}

Join BYJU'S Learning Program

If G1 and G2 are two geometric means and A is the arithmetic mean inserted between two positive numbers , then the value of G21G2+G22G1 is

The correct option is B 2ANow if two geometric means are inserted between a3 and b3 that means b3 is the fourth term of that GP.So a3r3=b3r=b/aG1=a3r=a2bG2=a3r2=ab2(G1)2/G2+(G2)2/G1=[a4b2/ab2]+[a2b4/a2b]=a3+b3=2A

If $G_{1}$ and $G_{2}$ are two geometric means and $A$ is the arithmetic mean inserted between two positive numbers , then the value of $\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}$ is