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Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers). $$0 . \overline{27}=0.272727 \ldots$$

Short Answer

Expert verified
Question: Convert the repeating decimal 0.272727... into a fraction. Solution: The repeating decimal 0.272727... can be converted into the fraction \(\frac{1}{3}\).

Step by step solution

01

Write the repeating decimal as a geometric series

To do this, we need to identify the first term and the common ratio of the geometric series. We observe that the repeating decimal can be written as: $$0.27 + 0.00027 + 0.00000027 + \ldots$$
02

Identify the first term and the common ratio

Here, the first term (a) is 0.27, and the common ratio (r) is 0.01 since we divide each consecutive term by 100 to obtain the next term in the sequence. We can write the geometric series as: $$0.27 + 0.27(0.01) + 0.27(0.01)^2 + \ldots$$
03

Use the formula for the sum of an infinite geometric series to find the fraction

The sum (S) of an infinite geometric series can be expressed as: $$S = \frac{a}{1-r}$$ where a is the first term and r is the common ratio. Plug in the values for a and r: $$S = \frac{0.27}{1-0.01}= \frac{0.27}{0.99}$$ Now, we can write the fraction as a ratio of two integers: $$\frac{27}{99}$$
04

Simplify the fraction

To simplify the fraction, we need to find the greatest common divisor (GCD) of the numerator and the denominator. In this case, the GCD is 27. So, we'll divide both the numerator and the denominator by 27: $$\frac{27\div27}{99\div27}=\frac{1}{3}$$ So, the repeating decimal \(0.\overline{27}\) can be expressed as the fraction \(\frac{1}{3}\).

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Key Concepts

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

Geometric Series
Picture a sequence of numbers where each term after the first is determined by multiplying the previous one by a consistent number. This sequence is known as a geometric series. Crucial to understanding these series are two terms: the first term, often denoted as 'a', and the common ratio, denoted as 'r'. This ratio is what we multiply by to get from one term to the next.

For example, the series 2, 4, 8, 16, ... is geometric because each term is twice (common ratio of 2) the previous one. In the context of repeating decimals, we can express them as geometric series by identifying the repeating block and treating it as the first term, and the positional shift as the common ratio. Illustrating this with the repeating decimal 0.2727..., we see the repeating block '27' which forms the series 0.27, 0.0027, 0.000027, and so on, where each term is one hundredth of the previous one, giving us a common ratio of 0.01.
Infinite Geometric Series Sum
With an infinite geometric series, the sequence of terms continues forever. Fascinatingly, under certain conditions, the sum of all these infinite terms can be a finite number. Specifically, when the common ratio 'r' is between -1 and 1, this sum converges, meaning it approaches a particular value without ever overshooting it.

The formula to calculate this sum is given by \( S = \frac{a}{1 - r} \), where 'S' is the sum, 'a' is the first term, and 'r' is the common ratio. In the case of the repeating decimal from our exercise, the series 0.27, 0.0027, 0.000027,... has a sum which can be calculated using this elegant formula. This is how the infinite sequence of a repeating decimal can be regarded as a neat, finite fraction.
Converting Decimals to Fractions
The process of converting decimals to fractions is a vital skill for making sense of decimal numbers beyond mere digits. Decimals that repeat infinitely often present a special case for conversion. We use the concept of infinite geometric series to view these repeating decimals as a sum, which can then be simplified into a fraction.

To convert a repeating decimal like \(0.\overline{27}\) to a fraction, we follow the geometric series method to arrive at an expression for the sum, and simplify to get the smallest integer values in the numerator and denominator. This simplification often involves finding the greatest common divisor (GCD) to reduce the fraction to its simplest form. This technique sheds light on the true value behind the curtain of perpetually cycling digits.

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Most popular questions from this chapter

James begins a savings plan in which he deposits \(\$ 100\) at the beginning of each month into an account that earns \(9 \%\) interest annually or, equivalently, \(0.75 \%\) per month. To be clear, on the first day of each month, the bank adds \(0.75 \%\) of the current balance as interest, and then James deposits \(\$ 100\). Let \(B_{n}\) be the balance in the account after the \(n\) th deposit, where \(B_{0}=\$ 0\). a. Write the first five terms of the sequence \(\left\\{B_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{B_{n}\right\\}\). c. How many months are needed to reach a balance of \(\$ 5000 ?\)

Suppose that you take 200 mg of an antibiotic every 6 hr. The half-life of the drug is 6 hr (the time it takes for half of the drug to be eliminated from your blood). Use infinite series to find the long-term (steady-state) amount of antibiotic in your blood.

Give an example of a bounded sequence that has a limit.

The famous Fibonacci sequence was proposed by Leonardo Pisano, also known as Fibonacci, in about \(\mathrm{A.D.} 1200\) as a model for the growth of rabbit populations. It is given by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1 .\) Each term of the sequence is the sum of its two predecessors. a. Write out the first ten terms of the sequence. b. Is the sequence bounded? c. Estimate or determine \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}},\) the ratio of the successive terms of the sequence. Provide evidence that \(\varphi=(1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Use induction to verify the remarkable result that $$f_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n}-(-1)^{n} \varphi^{-n}\right).$$

Estimate the value of the following convergent series with an absolute error less than \(10^{-3}\). $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2 k+1) !}$$

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