Question

find the 8th term of the geometric sequence, given the first term and common ratio. a1=6 r=-1/3

Answer 1

The rule of a geometric sequence:

$a_n=a_1r^{n-1}$

We have:

$a_1=6;\ r=-\dfrac{1}{3}$

Find $a_8=?\to n=8$

substitute:

$a_8=6\cdot\left(-\dfrac{1}{3}\right)^{8-1}=6\cdot\left(-\dfrac{1}{3}\right)^7=6\cdot\left(-\dfrac{1}{2187}\right)=-\dfrac{2}{729}$

Answer 2

The 8th term of the geometric sequence is -2/729 if the first term and common ratio. a1=6 r=-1/3, the answer is -2/729.

What is a sequence?

It is defined as the systematic way of representing the data that follows a certain rule of arithmetic.

We have:

A geometric sequence with the first term is 6 and the common difference is -1/3

As we know, the geometric sequence can represented by:

a, ar, ar², ar³,...

a is the first term and r is a common difference.

a₁ = 6

r = -1/3

The 8th term can be given as:

a₈ = a₁rⁿ⁻¹

n = 8, a₁ = 6, r = -1/3

a₈ = 6(-1/3)⁸⁻¹

a₈ = 6(-1/3)⁷¹

a₈ = -2/729

Thus, the 8th term of the geometric sequence is -2/729 if the first term and common ratio. a1=6 r=-1/3, the answer is -2/729.

Learn more about the sequence here:

brainly.com/question/21961097

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