Question

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

Answer 1

@ospreytriple @Mehek14

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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 \(t_4\).

Answer 6

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

Answer 7

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.

Answer 8

is t3=-3t3-1
t3=-3t2
t3=-3(-12)
t3=36

like that?

Answer 9

Excellent! Now, what's \(t_4\) ?

Answer 10

t4=-3t4-1
-3t3
-3(36)
-108

like that?

Answer 11

Congrats! Problem finished.

Answer 12

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

Answer 13

I agree. You're welcome.

Answer 14

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

Answer 15

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