Given the sequence 6,2, 23,29,…,26561; How many terms are there?

−26=−13 - this is the common ratio The sequence is: 6,−2,23,−29,227,−281,2243,−2729,22187,−26561, Number of terms =10⇒Total Sum=29,5246,561

First term a=6 Common ratio r=−23=−13 Number of terms n=10 Sum of 10 terms =a(r10−1)/(r−1)=6((−1/3)101)/(−1/3−1)=6×−3/4(−1/3)101)=−9/2(−1/3)101)

−26=−13 - this is the common ratio The sequence is: 6,−2,23,−29,227,−281,2243,−2729,22187,−26561, Number of terms =10⇒Total Sum=29,5246,561

First term a=6 Common ratio r=−23=−13 Number of terms n=10 Sum of 10 terms =a(r10−1)/(r−1)=6((−1/3)101)/(−1/3−1)=6×−3/4(−1/3)101)=−9/2(−1/3)101)

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High School
Algebra
Algebra I
Algebra foundations

Gregory Jones

Answered question

2021-12-28

Given the sequence 6,2,

\frac{2}{3}, \frac{2}{9}, \dots, \frac{2}{6561}

; How many terms are there?

Answer & Explanation

Mason Hall

Beginner2021-12-29Added 36 answers

Step 1
The first term

= a = 6

The common ratio

= r = \frac{second term}{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^{(n - 1)}

\frac{2}{6561} = 6 {(\frac{1}{3})}^{(n - 1)}

\frac{2}{6561} \cdot \frac{1}{6} = {(\frac{1}{3})}^{(n - 1)}

\frac{1}{19683} = {(\frac{1}{3})}^{(n - 1)}

{(\frac{1}{3})}^{9} = {(\frac{1}{3})}^{(n - 1)}

9 = n - 1

n = 10

Answer: 10

Nadine Salcido

Beginner2021-12-30Added 34 answers

- \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 \Rightarrow Total Sum = \frac{29, 524}{6, 561}

karton

Expert2022-01-04Added 613 answers

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

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