There are 10 terms in the geometric progression.

[nae99]

Homework Statement

3,6,12...1536
determine the number of terms in the progression

Homework Equations

The Attempt at a Solution

a=3 r=2

n= ar^n-1

1536= (3) (2)^n-1

 

[HallsofIvy]

Okay, so solve it! First, divide both sides by 3. If that sequence is actually geometric, you should be able to identify 1536/3 as a power of 2. I suggest you just calculate powers of 2: 1, 2, 4, 8, 16, ... until you get to that number.

 

[nae99]

Okay, so solve it! First, divide both sides by 3. If that sequence is actually geometric, you should be able to identify 1536/3 as a power of 2. I suggest you just calculate powers of 2: 1, 2, 4, 8, 16, ... until you get to that number.

1536= (3) (2)^n-1

1536/3 = 6^n-1$/$3

512 = 2^n-1

512 = 2^10-1

512 = 2^9

n = 10

 

[Mark44]

1536= (3) (2)^n-1

1536/3 = 6^n-1$/$3

512 = 2^n-1

512 = 2^10-1

512 = 2^9

n = 10

There are errors in your work. Also, you need parentheses around your exponent expressions.

1536= (3) (2)^(n-1)

1536/3 = 6^n-1$/$3

The above is incorrect. 3*2^(n - 1) $\neq$ 6^(n - 1)

512 = 2^n-1

The above is also incorrect. [6^(n - 1)]/3 $\neq$ 2^(n - 1)

512 = 2^10-1

512 = 2^9

n = 10