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
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
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.
Hello, sigma1!
$\displaystyle \text{Obtain the number of terms in the following geometric progression:}$
. . . . $\displaystyle 81,\;27,\;9,\;\hdots\;\dfrac{1}{243}$
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 are ten terms in the progression.