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Find the indicated term of each geometric sequence. 10th term of \(-1,2,-4, \ldots\)

Short Answer

Expert verified
The 10th term is 512.

Step by step solution

01

Identify the first term and common ratio

The first term of the geometric sequence is given as onumber\(a_1 = -1otitleotag\). We need to find the common ratio, which is the factor by which we multiply each term to get to the next term. Using the first two terms: onumber\(\frac{2}{-1} = -2otitleotag\), the common ratio (r) is onumber\(r = -2otitleotag\).
02

Use the formula for the n-th term of a geometric sequence

The formula for the n-th term of a geometric sequence is given by onumber\(a_n = a_1 \times r^{(n-1)}otitleotag\). Substituting the values we have: onumber\(a_{10} = -1 \times (-2)^{10-1}otitleotag\).
03

Calculate the 10th term

Now calculate onumber\((-2)^{9}otitleotag\). This equals onumber\(-512otitleotag\). Therefore, onumber\(a_{10} = -1 \times -512 = 512otitleotag\).
04

Write the final answer

The 10th term of the geometric sequence is onumber\(512otitleotag\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Term in Geometric Sequences
The first term of a geometric sequence is the initial value from which the sequence begins. It is often denoted as \(a_1\). In our exercise, the first term \(a_1\) is given as \(-1\). This term is crucial because it sets the stage for the entire sequence. Every subsequent term is derived based on this initial value.
Understanding the first term helps us to see where the sequence starts and makes it easier to work with the formula for finding other terms in the sequence.
Common Ratio in Geometric Sequences
The common ratio, often denoted by \(r\), is the factor by which each term of the sequence is multiplied to get the next term. It can be found by dividing the second term by the first term of the sequence. In our exercise, the first two terms are \(-1\) and \(2\). So, the common ratio is:
\(r = \frac{2}{-1} = -2\)
The common ratio is crucial because it determines how the sequence progresses. If the ratio is positive, the terms will stay either all positive or all negative. If the ratio is negative, like in our case, the signs of the terms will alternate. This alternating nature is evident from the sequence: \(-1, 2, -4, \text{...}\).
Finding the N-th Term of a Geometric Sequence
The formula for finding the n-th term of a geometric sequence is given by:
\(a_n = a_1 \times r^{(n-1)}\)
This formula lets us find any term in the sequence if we know the first term \(a_1\) and the common ratio \(r\). In our exercise, we are asked to find the 10th term. Substituting the values for \(a_1 = -1\) and \(r = -2\), we use:\
10th term formula calculation:
\(a_{10} = -1 \times (-2)^{10-1}\)
This simplifies to \(a_{10} = -1 \times (-2)^9\). We calculate \((-2)^9\) which equals \(-512\). So, the term \(a_{10}\) is:
\(a_{10} = -1 \times -512 = 512\)
Thus, the 10th term is \(512\). This formula is powerful because it allows us to compute any term in the sequence without having to list all previous terms.

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Most popular questions from this chapter

Bode's Law In \(1772,\) Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun: $$ a_{1}=0.4 \quad a_{n}=0.4+0.3 \cdot 2^{n-2} $$ where \(n \geq 2\) is the number of the planet from the sun. (a) Determine the first eight terms of the sequence. (b) At the time of Bode's publication, the known planets were Mercury \((0.39 \mathrm{AU}),\) Venus \((0.72 \mathrm{AU}),\) Earth \((1 \mathrm{AU})\) Mars \((1.52 \mathrm{AU}),\) Jupiter \((5.20 \mathrm{AU}),\) and Saturn \((9.54 \mathrm{AU})\) How do the actual distances compare to the terms of the sequence? (c) The planet Uranus was discovered in \(1781,\) and the asteroid Ceres was discovered in \(1801 .\) The mean orbital distances from the sun to Uranus and Ceres " are \(19.2 \mathrm{AU}\) and \(2.77 \mathrm{AU},\) respectively. How well do these values fit within the sequence? (d) Determine the ninth and tenth terms of Bode's sequence. (e) The planets Neptune and Pluto" were discovered in 1846 and \(1930,\) respectively. Their mean orbital distances from the sun are \(30.07 \mathrm{AU}\) and \(39.44 \mathrm{AU},\) respectively. How do these actual distances compare to the terms of the sequence? (f) On July \(29,2005,\) NASA announced the discovery of a dwarf planet \((n=11),\) which has been named Eris. Use Bode's Law to predict the mean orbital distance of Eris from the sun. Its actual mean distance is not yet known, but Eris is currently about 97 astronomical units from the sun.

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