obtain the number of terms in the following geometric progression

81, 27 , 9 , 1/243

can i find the number of terms using this

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

the answer that i get is a fraction though and that couldnt be possibly correct

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- Dec 8th 2010, 09:20 AMsigma1number 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

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

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

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

Quote:

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

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

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

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

So we have: .$\displaystyle 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: .$\displaystyle a_n \:=\:\dfrac{1}{243}$

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

Therefore, there areterms in the progression.*ten*

- Dec 8th 2010, 11:52 AMsigma1
- Dec 8th 2010, 12:55 PMemakarov