# Formula for the sequence

• September 22nd 2011, 04:57 PM
l flipboi l
Formula for the sequence

I know the difference is (-2/3), but I can't seem to figure out the formula.

Thanks,
• September 22nd 2011, 05:10 PM
TheChaz
Re: Formula for the sequence
The ratio is (-2/3)

the nth term is given by $(-8)* \cdot (-2/3)^{n-1}$
• September 22nd 2011, 05:14 PM
Plato
Re: Formula for the sequence
Quote:

Originally Posted by l flipboi l
Hello, can someone please help me find the formula for the sequence? I know the difference is (-2/3), but I can't seem to figure out the formula.

Try $a_n=\frac{(-2)^{n+3}}{3^{n}},~n=0,1,\cdots$
• September 22nd 2011, 05:25 PM
deepashree
Re: Formula for the sequence
Hi,
The nth term is given by Tn=a+(n-1)d
=-8+(n-1)(-2/3)
=-11/3(2+n).
• September 22nd 2011, 05:25 PM
l flipboi l
Re: Formula for the sequence
Quote:

Originally Posted by Plato
Try $a_n=\frac{(-2)^{n+3}}{3^{n}},~n=0,1,\cdots$

thx, but n is suppose to start at 1.

so the first term a1 = -8.
• September 22nd 2011, 05:26 PM
l flipboi l
Re: Formula for the sequence
thanks!
• September 22nd 2011, 05:36 PM
TheChaz
Re: Formula for the sequence
Quote:

Originally Posted by l flipboi l
thx, but n is suppose to start at 1.

so the first term a1 = -8.

That's an easy fix... (!)
Replace "n" with "n - 1" in P's solution.
• September 22nd 2011, 05:37 PM
skeeter
Re: Formula for the sequence
Quote:

Originally Posted by deepashree
Hi,
The nth term is given by Tn=a+(n-1)d
=-8+(n-1)(-2/3)
=-11/3(2+n).

fyi, the sequence is geometric, not arithmetic.
• September 22nd 2011, 07:54 PM
l flipboi l
Re: Formula for the sequence
Thanks, everyone!