-   -   How do you write the equation for this sequence 3, -6, 12, -24? (https://www.askmehelpdesk.com/showthread.php?t=432025)

• Jan 6, 2010, 02:52 AM
geelo
how do you write the equation for this sequence 3, -6, 12, -24?
I have an assignment and this is to find the equation of the sequence 3, -6, 12, -24? Please help me! Thanks
• Jan 6, 2010, 02:57 AM
Curlyben
What do you think the pattern is ?
• Jan 6, 2010, 03:25 AM
geelo
multiplying the preceding term by -2 to get the next term... but what is the equation?
• Jan 6, 2010, 03:28 AM
Curlyben
Well you've got it,
so
$x = ?$
• Jan 6, 2010, 10:20 AM
Unknown008

Well, the general way to write the 'general solution of a sequence' of this type (geometric progression) is that way:

$T_n = ar^{n-1}$

where n is the n th term (for example T1 is first term, T2 second term, etc)
a is the first term
r is the common ratio

Well, you have Equations setting up:
$T_1 = 3 = (3)(r^{1-1})$
$T_2 = -6 = (3)(r^{2-1})$
$T_3 = 12 = (3)(r^{3-1})$

Those are enough to give the formula:

$T_n = (3)(-2^{n-1})$

Anyway, that's how you do it in higher classes if ever I made a wrong guess on your math level.
• Jan 14, 2010, 03:39 AM
geelo
Quote:

Originally Posted by Unknown008
Well, the general way to write the 'general solution of a sequence' of this type (geometric progression) is that way:

$T_n = ar^{n-1}$

where n is the n th term (for example T1 is first term, T2 second term, etc)
a is the first term
r is the common ratio

Well, you have Equations setting up:
$T_1 = 3 = (3)(r^{1-1})$
$T_2 = -6 = (3)(r^{2-1})$
$T_3 = 12 = (3)(r^{3-1})$

Those are enough to give the formula:

$T_n = (3)(-2^{n-1})$

Anyway, that's how you do it in higher classes if ever I made a wrong guess on your math level.

thanks you very much for the equation...
have a nice day!
god bless! :)
• Jan 15, 2010, 12:02 PM
Unknown008

You're welcome! :)

God bless too :)

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