number of terms in geometric progression

**sigma1** · December 8th 2010, 09:20 AM

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

**emakarov** · December 8th 2010, 10:24 AM

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.

**Soroban** · December 8th 2010, 10:43 AM

Hello, sigma1!

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

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

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.

**sigma1** · December 8th 2010, 11:52 AM

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
$S_n = a(1-r^n) /1-r$

**emakarov** · December 8th 2010, 12:55 PM

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.

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