What Is The Next Number In The Sequence? 9….3….1….1/3…?

G.P. with a=9 & r=(1/3). So next 3 numbers in above sequence are 1/9, 1/27, 1/81.

What is the next number in the sequence 9 3 1 1 3?

Solution: Given, the sequence is 9.3.1.1/3. We have to find the next number in the sequence. From the sequence, We observe that the first number divided by 3 gives the second number, the second number divided by 3 gives the third number and so on.9/3 = 3 3/3 = 1 1/3 = 1/3 The next number will be 1/3 divided by 3.

What type of sequence is this 9 3 1?

Solution: The given sequence is a geometric progression, The first term is 9 and the second term is 3. It is obtained by dividing 9 by 3. Similarly, the third term(1) is obtained by dividing the second term(3) by 3. Down the line, the fifth term is obtained by dividing the fourth term by 3.

What is the common ratio of 27 9 3 1 1 3?

Solution: Given: Sequence is 27, 9, 3, 1, A sequence in which the ratio between two consecutive terms is the same is called a geometric sequence. A geometric progression is a sequence where every term bears a constant ratio to its preceding term. The geometric sequence is generally represented in form a, ar, ar 2,,

1. Where a is the first term and r is the common ratio of the sequence.
2. The c common ratio can have both negative as well as positive values.
3. The common ratio of a geometric progression is calculated by dividing two consecutive terms and simplifying it to the simplest form.
4. A\(_2\)/a\(_1\) = 9/27 = 1/3 a\(_3\)/a\(_2\) = 3/9 = 1/3 a\(_4\)/a\(_3\) = 1/3 In the sequence, the ratio 1/3 is the same and is called the common ratio.

Therefore, the common ratio for the given sequence is 1/3. Summary: The common ratio between successive terms in the sequence 27, 9, 3, 1,. is 1/3.

What is the next number in the sequence 3 4 6 9 13?

Solution: Sequence is defined as a list of numbers (or items) that exhibits a particular pattern. Given: 3, 4, 6, 9, 13, 18, 24, ? From the given sequence, we see that the pattern followed is: 3 + 1 = 4 4 + 2 = 6 6 + 3 = 9 9 + 4 = 13 13 + 5 = 18 18 + 6 = 24 Thus, we see that every resulting term in the given sequence is getting added to consecutive integers,

How many times does 1 3 go into 9?

We know we want the denominator to be 3 as that is stated in the question, so to get from 1 to 3 we times by 3, so by doing the same to the top, we get (9 x 3)/3, which is 27/3, so there are 27 thirds in 9.

What kind of sequence is 1 1 1 1 3 6 9 12?

This type of sequence is called an arithmetic sequence.

How do you find the next number in a sequence?

How to find the next term in an arithmetic sequence – Algebra 1 Possible Answers: None of the other answers Explanation : The question states that the sequence is arithmetic, which means we find the next number in the sequence by adding (or subtracting) a constant term. We know two of the values, separated by one unknown value. We know that is equally far from -1 and from 13; therefore is equal to half the distance between these two values. The distance between them can be found by adding the absolute values. Given the sequence below, what is the sum of the next three numbers in the sequence? Possible Answers: Correct answer: Explanation :

• By taking the difference between two adjacent numbers in the sequence, we can see that the common difference increases by one each time.
• Our next term will fit the equation, meaning that the next term must be,
• After, the next term will be, meaning that the next term must be,
• Finally, after, the next term will be, meaning that the next term must be
• The question asks for the sum of the next three terms, so now we need to add them together.

1. We have the following sequence
3. What is the value of ?

Possible Answers: Correct answer: Explanation : First, find a pattern in the sequence. You will notice that each time you move from one number to the very next one, it increases by 7. That is, the difference between one number and the next is 7. Therefore, we can add 7 to 36 and the result will be 43.

• The sequence follows the pattern for the equation:
• Therefore,

Find the next term in the following sequence. Possible Answers: Correct answer: Explanation : Determine what kind of sequence you have, i.e. whether the sequence changes by a constant difference or a constant ratio. You can test this by looking at pairs of numbers, but this sequence has a constant difference (arithmetic sequence).

2. Subtract the second term from the third term.
4. Subtract the third term from the fourth term.
6. To find the next value, add to the last given number.

Find the next term in the following arithmetic sequence: Possible Answers: Correct answer: Explanation : First, find the common difference for the sequence. Subtract the first term from the second term.

• Subtract the second term from the third term.
• To find the next value, add to the last given number.

Find the next term in the given arithmetic sequence: Possible Answers: Correct answer: Explanation : First, find the common difference for the sequence. Subtract the first term from the second term.

2. Subtract the second term from the third term.
4. To find the next value, add to the last given number.

Find the next term in the following arithmetic sequence: Possible Answers: Correct answer: Explanation : First, find the common difference for the sequence. Subtract the first term from the second term.

• Subtract the second term from the third term.
• To find the next value, add to the last given number.

Find the next term in the following arithmetic sequence: Possible Answers: Correct answer: Explanation : First, find the common difference for the sequence. Subtract the first term from the second term.

2. Subtract the second term from the third term.
4. Subtract the third term from the fourth term.
6. To find the next value, add to the last given number.

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What is the common ratio of the geometric sequence 9 3 1 1 3 1 9 1 27 1 81?

The common ratio= 1/3. S=1/3, 1/9, 1/27, 1/81,

What is the sequence 1 3 5 7 9 11 an example of?

Sequences with such patterns are called arithmetic sequences. In an arithmetic sequence, the difference between consecutive terms is always the same. For example, the sequence 3, 5, 7, 9 is arithmetic because the difference between consecutive terms is always two.

What is the common ratio of 9 3 1 1 3?

The sequence 9,3,1,1/3, is a geometric sequence with common ratio 1/3.

What is the sequence 1 3 1 9 1 27 an example of?

This is a geometric sequence since there is a common ratio between each term.

What pattern is 1 3 9 27?

The rule of the geometric sequence 1, 3, 9, 27, 81, 243, is 3n where n is the n-th term in the sequence.

What is the next number in sequence 1 1 2 4 3 9 4?

So the final sequence will be 1,1,2,4,3,9,4, 16,5,25. Was this answer helpful?

What is the missing number in the following sequence 1 3 3 6 7 9 12 21 1 point?

Given sequence: 1, 3, 3, 6, 7, 9, _, 12, 21. Hence, the correct answer is ‘ 13 ‘.

What is the next number in the sequence 1 2 3 5 7 11 13 _____?

Why Do Math?

• Number Theory Primes: Jumping Champions
• Leaping over the gaps between prime numbers
• MATHEMATICAL RECREATIONS by Ian Stewart
• Scientific American, December 2000

Mathematics is full of surprises. Who would have imagined, for instance, that something as straightforward as the natural numbers (1, 2, 3, 4,.) could give birth to anything so baffling as the prime numbers (2, 3, 5, 7, 11,,)? The pattern of natural numbers is obvious: no matter which one you pick, it’s easy to determine what the next one is.

You can’t say that for the primes. And yet it’s a simple step from natural numbers to primes. Just take the natural numbers that have no proper divisors. We know a lot about the primes, including some powerful formulas that provide good approximations when exact answers aren’t forthcoming. The Prime Number Theorem states that the number of primes less than x is approximately x /log x, where log denotes the natural logarithm.

So, for instance, we know that there are roughly 4.3 x 10 97 primes with less than 100 digits—but the exact number is a total mystery. Andrew Odlyzko of AT&T Labs, Michael Rubinstein of the University of Texas and Marek Wolf of the University of Wroclaw in Poland turned their attention to the gaps between successive primes in an article in Experimental Mathematics (Vol.8, No.2, 1999), where they addressed the following problem: What number is the most common gap between successive primes less than x ? This question was posed in the late 1970s by Harry Nelson of Lawrence Livermore National Laboratory.

Later on, John Horton Conway of Princeton University coined the phrase “jumping champions” to describe these numbers. The primes up to 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. The sequence of gaps—the differences between each prime and the next—is 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2 and 4.

The number 1 appears only once because all primes except for 2 are odd. The rest of the gaps are even numbers. In this sequence, 2 occurs six times, 4 occurs five times, and 6 occurs twice. So when x = 50, the most common gap is 2, and this number is therefore the jumping champion.

Sometimes several gaps are equally common. For instance, when x = 5 the gaps are 1 and 2, and each occurs once. For higher x, the sole jumping champion is 2 until we reach x = 101, when 2 and 4 are tied for the honor. After that, the jumping champion is either 2 or 4, or both, until x = 179, when 2, 4 and 6 are involved in a three-way tie.

At that point the challenge from 4 and 6 dies away, and 2 reigns supreme until x = 379, where 2 is tied with 6. Above x = 389 the jumping champion is mostly 6, occasionally tied with 2 or 4, or both. But when x ranges from 491 to 541, the jumping champion reverts to 4.

From x = 947 onward the sole jumping champion is 6, and a computer search shows that this continues up to at least x = 10 12, It seems reasonable to conclude that apart from some initial competition from 1, 2 and 4, the only long-term jumping champion is 6. But even a pattern that persists up to numbers in the trillions may well change as the numbers get still bigger.

And that’s where the surprise comes in. Odlyzko and his colleagues provided a persuasive argument that some-where near x = 1.7427 x 10 35 the jumping champion changes from 6 to 30. They also suggest that it changes again, to 210, near x = 10 425,

1. Except for 4, the conjectured jumping champions fit into an elegant pattern, which becomes obvious if we factor them into primes:
3. 2 = 2 6 = 2 x 3 30 = 2 x 3 x 5
4. 210 = 2 x 3 x 5 x 7

Each number is obtained by multiplying successive primes together. These numbers are called primorials—like factorials, but using primes—and the next few are 2,310, 30,030 and 510,510. In their article, Odlyzko and his co-authors proposed the Jumping Champions Conjecture: the jumping champions are precisely the primorials, together with 4.

• Here’s a brief explanation of their analysis.
• Anyone who looks at the sequence of primes notices that every so often two consecutive odd numbers are prime: 5 and 7, 11 and 13, 17 and 19.
• The Twin Prime Conjecture states that there are infinitely many such pairs.
• It is based on the idea that primes occur “at random” among the odd numbers, with a probability based on the Prime Number Theorem.

Of course, this sounds like nonsense—a number is either prime or not; there isn’t any probability involved—but it is reasonable nonsense for this kind of problem. According to a calculation of probabilities, there is no chance that the list of twin primes is finite.

What about three consecutive odd numbers being prime? There is only one example: 3, 5, 7. Given any three consecutive odd numbers, one must be a multiple of 3, and that number is therefore not prime unless it happens to equal 3. Yet the patterns p, p + 2, p + 6 and p, p + 4, p + 6 cannot be ruled out by such arguments, and they seem to be quite common.

For example, the first type of pattern occurs for 11, 13, 17 and again for 41, 43, 47. The second type of pattern occurs for 7, 11,13 and again for 37, 41, 43. About 80 years ago English mathematicians Godfrey Harold Hardy and John Edensor Littlewood analyzed patterns of this kind involving larger numbers of primes.

• Using the same kind of probabilistic calculation that I described for the twin primes, they deduced a precise formula for the number of sequences of primes with a given pattern of gaps.
• The formula is complicated, so I won’t show it here; see the article in Experimental Mathematics and the references therein.

From the Hardy-Littlewood work, Odlyzko and his colleagues extracted a formula for N (x, d ), which is the number of gaps between consecutive primes when the gap is of size 2 d and the primes are less than x, (We use 2 d rather than d because the size of the gap has to be even.) The formula is expected to be valid only when 2 d is large and x is much larger.

• The illustration at the left shows how log N ( x, d ) varies with 2 d for 13 values of x ranging from 2 20 to 2 44 (in this graph, log denotes a base 10 logarithm).
• Each plot line is approximately straight but has lots of bumps.
• A particularly prominent bump occurs at 2 d = 210, the conjectured jumping champion for very large x,

(It would look even more prominent if the logarithmic graphing didn’t flatten it out.) This kind of information suggests that the N ( x, d ) formula is not too wide off the mark. Now, if 2 d is going to be a jumping champion, the value of N ( x, d ) has to be pretty big.

• The best way to achieve this is if 2 d has many distinct prime factors.
• Also, 2 d should be as small as possible subject to this condition, so the most plausible choices for 2 d are the primorials.
• The known jumping champion 4 is presumably an exception.
• It occurs at a size where the N ( x, d ) formula isn’t a good approximation anyway.

The formula also lets us work out roughly when a given primorial takes over from the previous one as the new jumping champion. What’s left for recreational mathematicians to do? Prove the Jumping Champions Conjecture, of course—or disprove it. If you can’t do either, try searching for other interesting properties of the gaps between primes.

What is the answer to 9 3 1 ⁄ 3 1?

*For those that cannot watch, the correct answer is 1. The correct answer is found by grouping 1/3 with parentheses and following the order of operations, with division taking precedence. The correct answer is 1.

What is 9 and 1 3 as a decimal?

Solution: 9 1/3 as a decimal is 9.33 When you convert 9 1/3 (or 28/3) to a decimal, 9.33 is your answer.

What is the answer to 8 9 1 3?

8/9 – 1/3 = 59 ≅ 0.5555556 Spelled result in words is five ninths.

What is the sequence 1 1 1 2 3 5 7?

Fibonacci Numbers (Sequence): 1,1,2,3,5,8,13,21,34,55,89,144,233,377, Fn=Fn−2+Fn−1 where n≥2. Each term of the sequence, after the first two, is the sum of the two previous terms. This sequence of numbers was first created by Leonardo Fibonacci in 1202.

What type of sequence is 1 3 1 9 1 15 1 21?

List Of Triangular Numbers – 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120,136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431, and so on.

What is the sequence 1 1 1 2 3 5?

What is the Fibonacci sequence? The Fibonacci sequence is a famous group of numbers beginning with 0 and 1 in which each number is the sum of the two before it. It begins 0, 1, 1, 2, 3, 5, 8, 13, 21 and continues infinitely.

What is the common ratio of the geometric sequence 9 3 1 1 3 1 9 1 27 1 81?

The common ratio= 1/3. S=1/3, 1/9, 1/27, 1/81,

What is in the sequence 3 9?

Solution: Given, The sequence 3, 9, 27, 81,, This is a geometric sequence since there is a common ratio between each of them. In this case, multiplying the previous term in the sequence by 3 gives the next term. A n = a 1 r n – 1, Where, r = 3, a 1 = 3.

Form of geometric sequence = a 1 r n – 1, Substituting, we get, a 5 = 3(3) 5 – 1 = 3(3) 4 = 3(81) = 243 a 6 = 3 (3) 6 – 1 = 3(3) 5 = 3(243) = 729 a 7 = 3 (3) 7 – 1 = 3(3) 6 = 3(729) = 2187 Therefore, the next three terms are 243, 729 and 2187. Summary: The next three terms of the sequence 3, 9, 27, 81,,

are243, 729 and 2187.

Which term of the sequence 1 − 1 3 1 9 − 1 27 is − 1 729?

Answer: The 6th term in the geometric sequence 1/3, 1/9, 1/27,, is 1/729. Step-by-step explanation: Given, The geometric sequence: 1/3, 1/9, 1/27,, To find, In which term does the given geometric sequence contain 1/729? Concept, aₙ = a(r)ⁿ ⁻ ¹,(1) where aₙ = nth term of G.P n = number of terms in the G.P a = first term in the G.P r = common ratio Calculation, Now, aₙ = 1/729, a = 1/3, and r = = 1/3, n = n substituting in equation (1) we get: (1/729) = (1/3)(1/3)ⁿ ⁻ ¹ ⇒ (1/729) = (1/3)ⁿ ⇒ 3ⁿ = 729 ⇒ 3⁶ = 729 ⇒ n = 6 Therefore, the 6th term in the geometric sequence 1/3, 1/9, 1/27,, is 1/729. #SPJ3

How many terms are there in the sequence 3 6 9 1 1 1?

Thus, the given sequence contains 37 terms.