## anonymous one year ago List the first four terms of a geometric sequence with t1 = 4 and tn = -3tn-1. im lost...again

1. anonymous

@ospreytriple @Mehek14

2. campbell_st

so for the 2nd term substitute n = 2 t2 = -3 x t1 so t2 = -3 x 4 t2 = -12 and then repeat the process for n =3 and n = 4 hope it helps

3. anonymous

For $$t_2$$, put 2 in for $$n$$ in the given formula, i.e.$t_2 = -3\times t_{2-1} = -3\times t_1$and you know what $$t_1$$ is.

4. anonymous

i got -12, is there more i have to do?

5. anonymous

Correct. Now you have $$t_1$$ and $$t_2$$. The question asks for the first four terms, so you need to repeat what you did above and calculate $$t_3$$ and $$t_4$$.

6. anonymous

so for t3 i do t3=-12 x t3-1?

7. anonymous

Not quite. The formula is$t_n = -3 t_{n-1}$The -3 is a constant, it never changes. In fact, when you understand the pattern, you'll see that -3 is the common ratio.

8. anonymous

is t3=-3t3-1 t3=-3t2 t3=-3(-12) t3=36 like that?

9. anonymous

Excellent! Now, what's $$t_4$$ ?

10. anonymous

t4=-3t4-1 -3t3 -3(36) -108 like that?

11. anonymous

Congrats! Problem finished.

12. anonymous

thank you!. im glade you someone who shows me then just tells me the anwser

13. anonymous

I agree. You're welcome.

14. anonymous

just woundering, but are you good with ellipses and hyperbola?

15. anonymous

okay i have another question...ill start another one