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

Question

aum · Answer

Is this the correct problem?

$\large t_1 = 4$

$\large t_n = -3t_{n-1}$

anonymous · Answer

yep

aum · Answer

"n"  stands for the number of the term: 1, 2, 3, etc.
$t_n$ stands for the $n^{th}$ term.  $t_1$ is the first term, $t_2$ is the second term, etc.  
So if you want the second term, put n = 2 in $\large t_n = -3t_{n-1}$
What do you get?

anonymous · Answer

-3t2-1?

aum · Answer

$\large t_n = -3t_{n-1}$
put n = 2
$\large t_2 = -3t_{2-1} = -3 * t_1$
Since $\large t_1 = 4$,  $\large t_2 = -3 * 4 = -12$

put n = 3 in $\large t_n = -3t_{n-1}$  to find $\large t_3$.  What is the number you get for $\large t_3$ ?

aum · Answer

A quick way to solve this problem is to interpret what $\large t_n = -3 * t_{n-1}$ means.

It says, any term is -3 times the previous term.

So if first term is 4, the second term is -3 times the first term or -3 * 4 = -12.

The third term is -3 times the second term = -3 * -12 = 36

The fourth term is -3 times the third term = -3 * 36 = -108

So the first four terms of the sequence are: 4, -12, 36, -108

anonymous · Answer

oh okay!! thank you so much!

aum · Answer

You are welcome.