Chapter 12: Q 12.45 (page 1189)

Find the thirteenth term of a sequence where the first term is $81$ and the common ratio is $r = \frac{1}{3}$ .

Short Answer

Expert verified

The thirteenth term of a sequence is equal to $\frac{1}{19683}$ .

Step by step solution

Step 1. Given information

The first term is $81$ and the common ratio is $r = \frac{1}{3}$ .

Step 2. Find the thirteenth term of a sequence.

Let $a, r$ denote the first term and the common ratio respectively.

role="math" $\begin{array}{rcl} a & = & 81 \\ r & = & \frac{1}{3} \\ a_{n} & = & a r^{n - 1} \\ a_{13} & = & 81 {(\frac{1}{3})}^{13} \end{array}$

$= 3^{4} {(\frac{1}{3})}^{13} = \frac{1}{3^{9}} = \frac{1}{19683}$

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Find the thirteenth term of a sequence where the first term is $81$ and the common ratio is $r = \frac{1}{3}$ .

Short Answer

Step by step solution

Step 1. Given information

Step 2. Find the thirteenth term of a sequence.

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Find the thirteenth term of a sequence where the first term is 81and the common ratio is r=13.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Find the thirteenth term of a sequence.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Calculus

Applied Mathematics

Decision Maths

Geometry

Logic and Functions

Study anywhere. Anytime. Across all devices.

Find the thirteenth term of a sequence where the first term is $81$ and the common ratio is $r = \frac{1}{3}$ .