# Given the sequence 3, 6, 12, 24, ... calculate the 9th and the nth terms of the sequence .

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3,6,12,24,.....

We note that the constance term is r= 2

a1= 3

a2= 3*r = 3*2 = 6

A3= 3*r^2 = 3*2^2 = 3*4 = 12

a4 = 3*r^3 = 3*8 = 24

.....

a9= 3*r^8 = 3*2^8 = 3*256 = 768

an = 3^r^(n-1)= 3*2^(n-1)

First, we'll from ratios from 2 consecutive terms of the given sequence:

6/3 = 2

12/6 = 2

24/12 = 2

We notice that all quotients are the same, so, the sequence is a geometric progression, whose first terms is a1 = 3 and the common ratio is r = 2.

We'll calculate a9:

a9 = a1*r^(9-1)

a9 = 3 * 2^8

a9 =3*256

**a9 = 768**

The standard formula for any term of a geometric progression is:

**an = a1*r^(n-1)**