Question

` Insert two geometric means between `1/sqrt(2)&2
$1 \div \sqrt{2}$
and
$2$


​

Answer 1

Given

• 1st term of GP (a) = 1/√2

• Last term of GP(b) = 2

To find

• Insert 2 geometric mean between these 2 digits.

Solution

Let's assume that the two geometric means between 1/√2 and 2 are G1 and G2.

→ Here a = 1/√2 and b = 2 .

So the GP will be :-

→ 1/√2 , G1 , G2 , 2 with common ratio as r .

And we know a formula which is :-

$r = ( { \frac{b}{a} )}^{ \frac{1}{n + 1} }$

Here n refers to the number of geometric mean .

Using this formula over here .

→ r = [2/(1/√2)]^(1/2+1)

→ r = (2√2)^(⅓)

→ G1 = ar

$= \: \frac{1}{ \sqrt{2} } \times {2 \sqrt{2} }^{ \frac{1}{3} }$

→ (√2)^(-2)×√2^(⅓)×2^(⅓)

→ (√2)^(-2 + ⅓)× [(√2)²]^(⅓)

→ (√2)^(-5/6 + ⅔ )

→ (√2)^(-1/6)

G1 = ( √2 )^(-1/6)

Now similarly we'll find G2

→ G2 = ar²

→( √2)^(-2)×(√2)^(2/3)×(√2)^(4/3)

→ (√2)^(2/3+4/3-2)

→ (√2)^(-1/3)

So the value of G2 is (√2)^(-1/3)