# number of terms in geometric progression

• December 8th 2010, 09:20 AM
sigma1
number of terms in geometric progression
obtain the number of terms in the following geometric progression

81, 27 , 9 , 1/243

can i find the number of terms using this

$ar^{n-1} = 1/243$

the answer that i get is a fraction though and that couldnt be possibly correct
• December 8th 2010, 10:24 AM
emakarov
I suppose there are some skipped terms between 9 and 1/243.

So $a = 81 = 3^4$ and $r = 1/3$. The equation is $3^4(1/3)^{n-1}=1/243=3^{-5}$. From here, $(1/3)^{n-1}=3^{-9}$, or $3^{1-n}=3^{-9}$, so n = 10.
• December 8th 2010, 10:43 AM
Soroban
Hello, sigma1!

Quote:

$\text{Obtain the number of terms in the following geometric progression:}$

. . . . $81,\;27,\;9,\;\hdots\;\dfrac{1}{243}$

This geometric progression has: . $\begin{Bmatrix}\text{first term:} & a_1 &=& 81 \\ \text{common ratio:} & r &=& \frac{1}{3} \end{Bmatrix}$

$\text{The }n^{th}\text{ term is: }\:a_n \:=\:a_1r^{n-1}$

So we have: . $a_n \;=\; 81\left(\dfrac{1}{3}\right)^{n-1} \;=\;3^4\cdot\dfrac{1}{3^{n-1}} \;=\;\dfrac{1}{3^{n-5}}$

We want: . $a_n \:=\:\dfrac{1}{243}$

. . $\dfrac{1}{3^{n-5}} \:=\:\dfrac{1}{3^5} \quad\Rightarrow\quad n - 5 \:=\:5 \quad\Rightarrow\quad n \:=\:10$

Therefore, there are ten terms in the progression.

• December 8th 2010, 11:52 AM
sigma1
Quote:

Originally Posted by emakarov
I suppose there are some skipped terms between 9 and 1/243.

So $a = 81 = 3^4$ and $r = 1/3$. The equation is $3^4(1/3)^{n-1}=1/243=3^{-5}$. From here, $(1/3)^{n-1}=3^{-9}$, or $3^{1-n}=3^{-9}$, so n = 10.

thanks alot.. i see where i went wrong.. if you were suposed to find the sum of the series though would you use this formula
$S_n = a(1-r^n) /1-r$
• December 8th 2010, 12:55 PM
emakarov
Quote:

Originally Posted by sigma1
if you were suposed to find the sum of the series though would you use this formula
$S_n = a(1-r^n) /(1-r)$

Yes.