# FIND THE GEOMETRIC MEAN OF THE GIVEN EXTREMES

Geometric mean the special type of average.

How to find geometric mean of two numbers ?

Let us consider a and b.

To find geometric mean of a and b, use the formula

= √(a  b)

How to find geometric mean of three numbers ?

Let us consider a, b and c.

To find geometric mean of a, b and c, use the formula

∛(a  b   c)

Note :

we multiply the numbers altogether and take the nth root of the multiplied numbers, where n is the total number of data values

Problem 1 :

Find the geometric mean of the following numbers

a) 5 and 25

b) 7 and 63

c) -2 and -8

Solution :

a) 5 and 25

Geometric mean = √(x1 x2)

= √(5 25)

= √125

= √(25 × 5)

= 5√5

So, geometric mean is 5√5.

b) 7 and 63

Geometric mean = √(x1 x2)

= √(7 63)

= √441

= 21

So, geometric mean is 21.

c) -2 and -8

Geometric mean = √(x1 x2)

= √((-2) (-8))

= √16

= 4

So, geometric mean is 4.

Problem 2 :

Insert 3 numbers between 4 and 324 such that the resulting sequence in a G. P.

Solution :

Let the numbers be G1, G2, and G3.

4, G1, G2, G3, 324 are in G.P.

T5 = 324

tn = ar(n  - 1)

ar4 = 324

4 × r4 = 324

Dividing 4 on both sides.

(4 × r4)/4 = 324/4

r4 = 81

r4 = 34

r = 3

G1 = 4 × 3 = 12

G2 = 4 × 32 = 36

G3 = 4 × 33 = 108

So, the numbers be 12, 36, 108.

Problem 3 :

A G.P is given 1/2, 1, 2, 4, … If a constant k is multiplied in each term of the G.P., then find the 10th term of the G.P.

Solution :

Accordingly the properties of geometric progression, if every term is multiplied by some non zero constant, then the new progression created is also a geometric progression with same common ratio.

To find the10th term of the G.P :

.an = a1rn  - 1

a10 = 1/2 × (2)10  - 1

= 1/2 × 29

= 1/2 × 256

a10 = 128

So, 10th term of the G.P is a10 = 128.

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