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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.