Gregory Jones

2021-12-28

Given the sequence 6,2, $\frac{2}{3},\frac{2}{9},\dots ,\frac{2}{6561}$; How many terms are there?

Mason Hall

Step 1
The first term $=a=6$
The common ratio $=r=\frac{\text{second term}}{\text{first term}}=\frac{2}{6}=\frac{1}{3}$
Let there are n terms.
Step 2
Using the term formula we get:
${a}_{n}=a{r}^{\left(n-1\right)}$
$\frac{2}{6561}=6{\left(\frac{1}{3}\right)}^{\left(n-1\right)}$
$\frac{2}{6561}\cdot \frac{1}{6}={\left(\frac{1}{3}\right)}^{\left(n-1\right)}$
$\frac{1}{19683}={\left(\frac{1}{3}\right)}^{\left(n-1\right)}$
${\left(\frac{1}{3}\right)}^{9}={\left(\frac{1}{3}\right)}^{\left(n-1\right)}$
$9=n-1$
$n=10$

$-\frac{2}{6}=-\frac{1}{3}$ - this is the common ratio
The sequence is:
$6,-2,\frac{2}{3},-\frac{2}{9},\frac{2}{27},-\frac{2}{81},\frac{2}{243},-\frac{2}{729},\frac{2}{2187},-\frac{2}{6561}$, Number of terms $=10⇒\text{Total Sum}=\frac{29,524}{6,561}$

karton

First term a=6
Common ratio $r=-\frac{2}{3}=-\frac{1}{3}$
Number of terms n=10
Sum of 10 terms
$\begin{array}{}\\ =a\left({r}^{10}-1\right)/\left(r-1\right)\\ =6\left(\left(-1/3{\right)}^{10}1\right)/\left(-1/3-1\right)\\ =6×-3/4\left(-1/3{\right)}^{10}1\right)\\ =-9/2\left(-1/3{\right)}^{10}1\right)\end{array}$

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