1.5.1 If all the terms in a Geometric Progression (GP) are multiplied or divided by the same quantity, the resulting terms will form a GP with the same common ratio as before.

1.5.2 If a, b, c, d are in Geometric Progression (GP) with a common ratio â€˜râ€™, they are also in continued proportion, as a/b = b/c = c/d = 1/r

1.5.3 If we need to consider 3 consecutive terms in a Geometric Progression (GP), we take it as a/r, a , ar

1.5.4 Four consecutive terms in a Geometric Progression(GP) are taken as (a/r^3 ), (a/r), (ar), (ar3)

1.5.5 Geometric Mean (GM): Geometric mean of n numbers a_{1}, a_{2},â€¦..a_{n} is given by (a_{1}Ã— a_{2}Ã— a_{3}Ã— â€¦â€¦a_{n} )^{(1/n)}

e.g. geometric mean of four numbers a, b ,c, d = (a b c d)^{1/4}

e.g. geometric mean of 2, 4, 8 = (2 Ã— 4 Ã— 8)^{1/3} = (64)^{1/3} = 4

If a, b, c are in GP, geometric mean, b = vac

i.e In a GP if you take three consecutive terms middle term will be geometric mean of first term and the third term.

And if you take a GP with 5 terms, them the middle term or 3rd term will be the GM of the series.

Application of GM :

The Concept of GM is used in finding out the compounded annual growth rate (CAGR)

CAGR = [((Final Value)/(Initial Value))^{(1/n)}-1]Ã— 100 where n is the number of years.

21. Population of a village is 10000 in year 2000 and is 14400 in 2002. Find CAGR?

CAGR = [(14400/10000)^{(1/2)}-1]Ã— 100 = 20 %

Meaning an annual increase of 20 %.

To verify

10000 Ã— 1.2 = 12000

12000 Ã— 1.2 = 14400

We are using a factor of multiplication 1.2 because to increase a number by 20% and to get the final value,

we need multiply by 120/100= 1.2

CAGR is similar to rate percent, R in compound interest calculations.

Final amount = P( 1+R/100)^{n} where P = initial amount, R = rate percent, n = number of years.

So, CAGR or R = [((Final Amount)/(Initial Amount))^{(1/n)}-1]Ã— 100

CAGR implies the annual increase in growth.

Inserting n GMâ€™s between A and B

When we insert n GMs between A and B, we will get a GP with n + 2 terms, the common ratio is given by

r = (b/a)^{(1/((n+1) ))}

GM_{1} = A.r

GM_{2} = A r^{2} â€¦â€¦â€¦â€¦

For example 2 GMs between 1 and 27 will be 3,9. We can find out the GMs by computing the common ratio=(27/1)^{(1/3)}= 3

We will get a GP with 4 terms 1, 3, 9, 27

3 GMs between between 1 and 16 will be 2, 4, 8 , so we will get a GP with 5 terms 1, 2, 4, 8, 16.

### Illustration:

22. Insert 8 geometric means between 3 and 1536.Find the 6th GM ?

Common ratio = (1536/3)^{(1/9)} = 2

GM_{1}= 3 Ã— 2 = 6

GM_{2}= 3 Ã— 2^{2} =12

GM_{3}= 3 Ã— 2^{3}= 24

â€¦â€¦

GM_{6}= 3 Ã— 2^{6}= 192 Required answer = 192. 23.

Find the number of GMs in a series where the ratio between the first and last terms is 16:1 and the common ratio of the Geometric Progression=2

Common ratio = (b/a)^{(1/((n+1) ))} . 2 = 16^{(1/x)}

2^{x} = 16 = > x = 4

x = n + 1= > n = 4 â€“ 1 = 3

Number of Geometric Means = 3