Write a formula for the general or nthn^{\operatorname{\mathrm{th}}} term for each geometric sequence.12,−6,3,−32,…12,-6,3,-\frac32,\ldots

a n = 12 ( − 1 2 ) n − 1 a_{n}=12\left(-\frac12\right)^{n-1}

Write a formula for the general or n t h n^{\operatorname{\mathrm{th}}} term for each geometric sequence. 12 , − 6 , 3 , − 3 2 , … 12,-6,3,-\frac32,\ldots

this problem, we need to find the formula for the general nth term of the geometric sequence 12, negative six, three, negative three halves. Now, to find this formula, we need two pieces of information. We need value of the first term of the sequence, and we need to know the common ratio. Now, we already have the value of the first term of our sequence. It's 12. So we'll have a sub n is equal to 12. But to find the common ratio, we can take one of our terms, like negative six, and divide it by the previous term, 12. Simplifying, we get that our common ratio is negative one half. So the formula for the general nth term is going to be 12 times negative one half to the n minus one power.

Write a formula for the general or n t h ⁡ n^{\operatorname{\mathrm{th}}} term for each geometric sequence. 12 , − 6 , 3 , − 3 2 , … 12,-6,3,-\frac32,\ldots

a n = 12 ( − 1 2 ) n − 1 a_{n}=12\left(-\frac12\right)^{n-1}

a n = 12 ( − 2 ) n − 1 a_{n}=12\left(-2\right)^{n-1}

a n = 12 ( 1 2 ) n − 1 a_{n}=12\left(\frac12\right)^{n-1} a n = 12 ( 2 1 ) n − 1

a n = ( − 1 2 ) n − 1 a_{n}=\left(-\frac12\right)^{n-1} a n = ( − 2 1 ) n − 1

15. Sequences, Series, and the Binomial Theorem

Geometric Sequences

Beginning & Intermediate Algebra

15. Sequences, Series, and the Binomial Theorem

Geometric Sequences

Struggling with Beginning & Intermediate Algebra?

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Multiple Choice

Write a formula for the general or $n^{t h}$ term for each geometric sequence.
$12, - 6, 3, - \frac{3}{2}, \dots$

$a_{n} = 12 {(- 2)}^{n - 1}$

$a_{n} = 12 {(- \frac{1}{2})}^{n - 1}$

$a_{n}=12$\left$($\frac$12$\right$)^{n-1}$

$a_{n}=$\left$(-$\frac$12$\right$)^{n-1}$

Verified step by step guidance

Identify the first term of the geometric sequence, which is the initial value given. In this sequence, the first term $a_1$ is 12.

Find the common ratio $r$ by dividing the second term by the first term. Calculate $r = \frac{a_2}{a_1} = \frac{-6}{12}$.

Simplify the common ratio to its simplest fractional form, which will include the sign to reflect the alternating positive and negative terms.

Write the general formula for the $n^{\operatorname{th}}$ term of a geometric sequence using the first term and common ratio: $a_n = a_1 \times r^{n-1}$.

Substitute the values of $a_1$ and $r$ into the formula to express the $n^{\operatorname{th}}$ term explicitly as $a_n = 12 \left(-\frac{1}{2}\right)^{n-1}$.

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Struggling with Beginning & Intermediate Algebra?

Write a formula for the general or nthn^{\(\operatorname{\mathrm{th}\)}} term for each geometric sequence.12,−6,3,−32,…12,-6,3,-\(\frac\)32,\(\ldots\)

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Write a formula for the general or $n^{t h}$ term for each geometric sequence.
$12, - 6, 3, - \frac{3}{2}, \dots$