# Geometric Sequences

1. Jul 27, 2009

1. The problem statement, all variables and given/known data

Hi, I was trying to work out this question, but i kinda got stuck. Can anyone help me please?
Thanks

4. Find the number of terms in each of these geometric sequences.

2,10,50....1250

2. Relevant equations

ar^n-1

3. The attempt at a solution

1250=2x5^n-1

2. Jul 27, 2009

### Cyosis

I take it you mean $1250=2*5^{n-1}$ with n starting at 1? You're on the right track. What is the problem with the expression you found so far? You are unable to solve it for n? Hint: logarithm

Last edited: Jul 27, 2009
3. Jul 27, 2009

i don't get what the next step is

4. Jul 27, 2009

### Cyosis

What does n represent in that formula?

5. Jul 27, 2009

n is the term, but i dont know how to find it

6. Jul 27, 2009

### Cyosis

"The term", you mean is the number of terms in the sequence. The first step in solving it is to put 5^(n-1) on one side and the other terms on the other side. Then take the logarithm on both sides.

7. Jul 27, 2009

so do i have to balance both sides and eliminate 1250?

8. Jul 27, 2009

### Cyosis

An equation is balanced by definition. If you substitute x=5^(n-1) then the equation becomes 1250=2x. Now solve this equation for x and then back substitute x.

ps. Are you familiar with logarithms?

9. Jul 27, 2009

Thank you very much. I've just started learning about it, so I'm trying to get the hang of it, but thank you for helping me.

10. Jul 27, 2009

### Cyosis

You're welcome, but have you found the solution?

11. Jul 27, 2009

1250= 2x so x=1250/2 x=625

oh no, I'm getting confused again. sorry

12. Jul 27, 2009

### Cyosis

Correct x=625 and because we substituted x=5^(n-1) we know that 5^(n-1)=625. We now want to write 625 in terms of 5 raised to a certain exponent. For example we can write 8=2^3, 125=5^3. Try to write 625 in a similar way.

13. Jul 27, 2009

so it can be 625=25^2

14. Jul 27, 2009

### Cyosis

Yes that is correct although not entirely what I had in mind. Note that you can write 25=5^2 therefore 25^2=(5^2)^2=5^4=625.

$$5^{n-1}=5^4$$

Therefore n-1 equals...?

15. Jul 27, 2009

n-1= 5

16. Jul 27, 2009

### Cyosis

No the two exponents need to be equal to each other, it is the only way the equation can hold. So I will ask you again n-1=?

What you're saying now is that 5^5=5^4. Convince yourself that this cannot be correct.

17. Jul 27, 2009

ohhhh n-1= 5-1?

18. Jul 27, 2009

### Cyosis

Why do you think so, explain your reasoning behind it.

19. Jul 27, 2009