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1. Find the geometric mean between:

a) 3 and 12
Solution:

GM = $\sqrt{3×12}$
$= \sqrt{36}$
$= 6$

c) 25 and 625
Solution:

GM = $\sqrt{25×625}$
$= \sqrt{15625}$
$= 125$

e) 8 and 32/3
Solution:

GM = $\sqrt{8 × \frac{32}{3}}$

$= \sqrt{\frac{256}{3}}$
$= \frac{16}{\sqrt{3}}$

2 a) Find the 2nd term of a geometric sequence whose first and third terms are 2/3 and 3/2 respectively.

Solution:
Here,
a = $\frac{2}{3}$
b = $\frac{3}{2}$
Now,
2nd term = GM between a and b
We know,
GM = $\sqrt{a × b}$
$= \sqrt{ \frac{2}{3} × \frac{3}{2}}$

$= \sqrt{1}$
$= 1$
Hence, the second term in between the given two terms is 1.


3 a) Two numbers are in the ratio of 2:1 and their geometric mean is 4. Find the numbers.

Solution:
Here,
$\dfrac{a}{b} = \dfrac{2}{1}$

$or, a = 2b$
Given,
$G = 4 = \sqrt{ab}$
$or, 4 = \sqrt{2b × b}$
Squaring both sides

$or, 4^2 = ( \sqrt{2b^2})^2$
$or, 16 = 2b^2$
$or, b^2 = 8$
$or, b = \sqrt{8}$
$\therefore b = 2\sqrt{2}$
And,

$a = 2× b = 2×2\sqrt{2} = 4\sqrt{2}$

Hence, the required numbers are 4√2 and 2√2.