Geometric Equations

[HNO]

Find the Sum of the geometric series whose 1st term is 9 and ratio is 2/3.

So I know the formula which I can use to find the sum are…



sn = t1 (1 - r ^n) / 1 - r

or

Sn= (rtn - t1) / (r-1)


But I don not know tn …so how am I suppose to solve for this? Unless this is the kind of question the involves in infinite series which I use this formula:

t1/ 1-r = s(infinity)*** That be way to easy right? 8.3333 or 25/3

otherwise if thats way wrong and I am suppose to use the geometric formula

WHERE DO I GO FROM HERE

Sn = 9 (1 - 2/3 ^n) / 1 - 2/3

***********************************************PLEASE HELP

 

[tkhunny]

This is the problem with a formula, one sometimes forgets WHEN to use it.

If you have r^n, you have a FINITE series.
If you haven't a finite series, get rid of that term.

 

[HNO]

This is the problem with a formula, one sometimes forgets WHEN to use it.

If you have r^n, you have a FINITE series.
If you haven't a finite series, get rid of that term.

\

So basically what you are saying is that it is an infinite series so is my answer right ?

 

[soroban]

Hello, HNO!

Find the sum of the geometric series whose 1st term is 9 and ratio is 2/3.

If we have a finite number of terms, \(\displaystyle n\)
. . the sum of the \(\displaystyle n\) terms is: .\(\displaystyle S_n \;=\;t_1\dfrac{1-r^n}{1-r}\)

If \(\displaystyle |r| < 1\), the infinite series has a finite sum:.\(\displaystyle S \;=\;\dfrac{t_1}{1-r}\)


Since they did not specify the number of terms,
. . we assume they want the sum of the infinite series.

Therefore: .\(\displaystyle S \;=\;\dfrac{9}{1-\frac{2}{3}} \;=\;\dfrac{9}{\frac{1}{3}} \;=\;27\)

 

[HallsofIvy]

It is not a matter of "assuming" anything. The opening sentence said "Find the Sum of the geometric series". A "series" is, by definition, an infinite sum.