Question

12 Find the geometric mean between :
(ii) 4/9 and 9/4
(i) 14 and 7/32
(iii) 2a and 8a³

Answer 1

Answer:

geometric mean =ab whole root

I...4/9×9/4whole root

=1

ii....32/7×14 root

64 root

=8

iii....2a×8a^3root

16a^4root

4a^2

Answer 2

i) The geometric mean of $\dfrac{4}{9}\ and \ \dfrac{9}{4}$ is 1.

ii) The geometric mean of $14\ and \ \dfrac{7}{32}=\dfrac{7}{4}$ .

iii) The geometric mean of 2a and 8a³ is 4a².

Explanation:

The geometric mean of two numbers a and b is given by :-

$GM=\sqrt{ab}$

i) The geometric mean of $\dfrac{4}{9}\ and \ \dfrac{9}{4}$ will be :

$GM=\sqrt{\dfrac{4}{9}\times \dfrac{9}{4}}$

$=\sqrt{1}=1$

So , the geometric mean of $\dfrac{4}{9}\ and \ \dfrac{9}{4}$  is 1.

ii) The geometric mean of $14\ and \ \dfrac{7}{32}$ will be :

$GM=\sqrt{14\times \dfrac{7}{32}}$

$=\sqrt{\dfrac{49}{16}}=\dfrac{\sqrt{49}}{\sqrt{16}}=\dfrac{7}{4}$

So , the geometric mean of $14\ and \ \dfrac{7}{32}=\dfrac{7}{4}$

iii) The geometric mean of 2a and 8a³ will be :

$GM=\sqrt{2a \times8a^3}$

$=\sqrt{16a^{1+3}}=\sqrt{16 a^4}$

$=\sqrt{16}\times\sqrt{(a^2)^2}$

$=4a^2$

Hence, the geometric mean of 2a and 8a³ is 4a².

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