## What Is The Nth Term?

### What is nth term in math?

• A is a list of numbers or diagrams that are in a particular order.
• For example, a number pattern which increases (or decreases) by the same amount each time is called an,
• The amount it increases or decreases by is known as the common difference,
• Recognising the common difference means that the sequence can be continued using a rule.
• The 𝒏th term refers to a term’s position in the sequence, for example, the first term has 𝒏 = 1, the second term has 𝒏 = 2 and so on. An expression for the 𝒏th term is worked out by looking at the difference between the terms of the sequence and comparing the sequence to the appropriate times table.

Between each pair of terms in this sequence is the amount the sequence is decreasing by. Watch the video to learn about arithmetic sequences and how any number in a sequence can be found using the \(n\) th term rule. An arithmetic sequence is a series of numbers that go up or down by the same amount each time.

In the song Ten Green Bottles Standing on the Wall, the sequence goes ten, nine, eight, seven, six and so on, reducing by one each time. This is called the common difference. In this sequence the common difference is -1 A term-to-term rule can be created to explain the common difference. In the sequence 11, 14, 17, 20, 23, it’s plus three because the sequence goes up by three with each term.

It might be necessary to work out a number much later in the sequence though. For example, what the 93rd number is and it would take ages to add three 92 times. This is where the 𝒏th term rule comes in handy. The 𝒏th term is a way of using the position of a term in a sequence, to find out what that term would be.

• Since the common difference is three, start by comparing the sequence to the three times table.
• Subtracting the number in the times table from the term results in a constant difference of plus eight between the table and the sequence.3 + 8 is 11, 6 + 8 is 14 and so on.
• To find the 93rd term multiply the common difference, 3 by 93, and then add 8 to get the answer, 287 The 𝒏th term rule can be expressed as 3𝒏 + 8, with 𝒏 representing the term needing to be found.

Sequences come up all the time in real life, so the 𝒏th term rule can really come in handy.

#### What is the nth term of 2 4 6 8?

Number Sequences – Maths GCSE Revision In the sequence 2, 4, 6, 8, 10. there is an obvious pattern. Such sequences can be expressed in terms of the nth term of the sequence. In this case, the nth term = 2n. To find the 1st term, put n = 1 into the formula, to find the 4th term, replace the n’s by 4’s: 4th term = 2 × 4 = 8. Example What is the nth term of the sequence 2, 5, 10, 17, 26. ?

• To find the answer, we experiment by considering some possibilities for the nth term and seeing how far away we are:
• n = 1 2 3 4 5
• n² = 1 4 9 16 25
• n²+1 = 2 5 10 17 26
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This is the required sequence, so the nth term is n² + 1. There is no easy way of working out the nth term of a sequence, other than to try different possibilities. Tips : if the sequence is going up in threes (e.g.3, 6, 9, 12.), there will probably be a three in the formula, etc.

1. n = 1 2 3 4 5
2. n(n + 1)/2 = 1 3 6 10 15
3. Clearly the required sequence is double the one we have found the nth term for, therefore the nth term of the required sequence is 2n(n+1)/2 = n(n + 1).
4. The Fibonacci sequence

The Fibonacci sequence is an important sequence which is as follows: 1, 1, 2, 3, 5, 8, 13, 21,,, The next term of this well-known sequence is found by adding together the two previous terms. : Number Sequences – Maths GCSE Revision

#### What is the nth term of 2 6 10 14?

The general (nth) term for 2, 6, 10, 14, 18, 22, is 4 and the first term is 2. If we let d=4 this becomes a n =a 1 +(n−1)d. The nth or general term of an arithmetic sequence is given by a n =a 1 +(n−1)d.

### What is the nth term of 1 2 4 8 16?

What is the nth term?

Type of Sequence Example n n n nth Term
Geometric 1,2,4,8,16,32, 1, 2, 4, 8, 16, 32,.1, 2, 4, 8, 16, 32, 1,2,4,8,16,32, 2n−1 2 n − 1 2 2n−1
Quadratic 3,9,19,33,51, 3, 9, 19, 33, 51,.3, 9, 19, 33, 51, 3,9,19,33,51, 2n2+1 2 n 2 + 1 2n^ +1 2n2+1

## What is the nth term of 15 12 9 6?

Solution: An arithmetic progression is a sequence where the difference between every two consecutive terms is the same. For a given arithmetic sequence, the nth term of AP is calculated using the following expression: a n = a + (n – 1) d Where,

‘a’ is the first term of the AP ‘d’ is the common difference ‘n’ is the number of terms ‘a n ‘ is the n th term of the AP.

Let’s find the nth term of the sequence, 15, 12, 9, 6, Here, a = 15, d = -3, n = n Thus, substituting these values in the formula a n = a + (n – 1) d ⇒ a n = 15 + (n – 1) (-3) ⇒ a n = 15 + 3 – 3n ⇒ a n = 18 – 3n Thus, the expression for the n th term of the sequence, 15, 12, 9, 6, is a n = 18 – 3n.

## What is the nth term of the sequence 1 3 5 7 9?

∴nth term is 2n−1.

### What is the nth term for 1 3 6 10?

Answer and Explanation: 1, 3, 6, 10,. Therefore, the n t h term = n ( n + 1 ) 2.

#### What is the nth term of 2 5 8 11 14?

The nth terms: 2,5,8,11,14, 17,20,23

### What is the nth term rule of 5 2 1 4 7?

What is the nth term of the sequence 5,2, -1, -4, -7? 8-3n is n’th term for this sequence. It’s an Arithmetic progression.

#### What is the nth term of 7 10 13 16 19?

The nth terms: 7,10,13,16,19, 22,25,28,31,34,37

## What is the nth term of 2 4 6 8 10 12?

Hence, the nth term of given numbers is 2n.

### What is the nth term of 5 9 13 17 21?

An arithmetic sequence (or arithmetic progression ) is a sequence (finite or infinite list) of real numbers for which each term is the previous term plus a constant (called the common difference ). For example, starting with 1 and using a common difference of 4 we get the finite arithmetic sequence: 1, 5, 9, 13, 17, 21; and also the infinite sequence 1, 5, 9, 13, 17, 21, 25, 29,,

• 4 n +1,,
• In general, the terms of an arithmetic sequence with the first term a 0 and common difference d, have the form a n = dn + a 0 ( n =0,1,2,.).
• If a 0 and d are relatively prime positive integers, then the corresponding infinite sequence contains infinitely many primes (see Dirichlet’s theorem on primes in arithmetic progressions).

An important example of this is the following two arithmetic sequences: 1, 7, 13, 19, 25, 31, 37,,5, 11, 17, 23, 29, 35, 41,, Together these two sequences contain all of the primes except 2 and 3. A related question is how long of a arithmetic sequence can we find all of whose members are prime.

1. Dickson’s conjecture says the answer should be arbitrarily long-but finding long sequences of primes is quite difficult.
2. It is fairly easy to heuristically estimate how many such primes sequences there should be for any given length-Hardy and Littlewood first did this in 1922,
3. In 1939, van der Corput showed that there are infinitely many triples of primes in arithmetic progression,
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Finally, in 2004, Green and Tao showed that there are indeed arbitrarily long sequences of primes and that a k -term one occurs before : 2 2 2 2 2 2 2 2 100 k Obviously this is not optimal! It is conjectured that it actually occurs before k !+1, The longest known arithmetic sequence of primes is currently of length 25, starting with the prime 6171054912832631 and continuing with common difference 366384*23#*n, found by Chermoni Raanan and Jaroslaw Wroblewski in May 2008.

• The longest known sequence of consecutive primes in arithmetic progression is ten starting with the 93-digit prime 100 9969724697 1424763778 6655587969 8403295093 2468919004 1803603417 7589043417 0334888215 9067229719, and continuing with common difference 210.
• See Tony Forbes’ web page for more information.) In August 2000 David Broadhurst found the smallest arithmetic progression of titanic primes of length three: 10 999 +61971, 10 999 +91737, 10 999 +121503; and of length four: 10 999 +2059323, 10 999 +2139213, 10 999 +2219103, 10 999 +2298993.

Jens Kruse Andersen’s excellent Primes in Arithmetic Progression Records Ten consecutive primes in arithmetic progression by Tony Forbes

References: Chowla44 S. Chowla, “There exists an infinity of 3-combinations of primes in A.P.,” Proc. Lahore Phil. Soc., 6 (1944) 15-16. MR 7,243l Corput1939 A.G. van der Corput, “Über Summen von Primzahlen und Primzahlquadraten,” Math. Ann., 116 (1939) 1-50.

DFLMNZ1998 H. Dubner, T. Forbes, N. Lygeros, M. Mizony, H. Nelson and P. Zimmermann, “Ten consecutive primes in arithmetic progression,” Math. Comp., 71 :239 (2002) 1323-1328 (electronic). MR 1 898 760 ( Abstract available ) DN97 H. Dubner and H. Nelson, “Seven consecutive primes in arithmetic progression,” Math.

Comp., 66 (1997) 1743-1749. MR 98a:11122 ( Abstract available ) GT2004a Green, Ben and Tao, Terence, “The primes contain arbitrarily long arithmetic progressions,” Ann. of Math. (2), 167 :2 (2008) 481-547. ( http://dx.doi.org/10.4007/annals.2008.167.481 ) MR 2415379 GT2004b B.

Green and T. Tao, “A bound for progressions of length k in the primes,” (2004) Available from http://people.maths.ox.ac.uk/greenbj/papers/back-of-an-envelope.pdf, Guy94 (A6) R.K. Guy, Unsolved problems in number theory, Springer-Verlag, 1994. New York, NY, ISBN 0-387-94289-0. MR 96e:11002 HL23 G.H. Hardy and J.E.

Littlewood, “Some problems of `partitio numerorum’ : III: on the expression of a number as a sum of primes,” Acta Math., 44 (1923) 1-70. Reprinted in “Collected Papers of G.H. Hardy,” Vol. I, pp.561-630, Clarendon Press, Oxford, 1966. Kra2005 B. Kra, “The Green-Tao theorem on arithmetic progressions in the primes: an ergodic point of view,” Bull.

Amer. Math. Soc., 43 :1 (2006) 3-23 (electronic). ( http://dx.doi.org/10.1090/S0273-0979-05-01086-4 ) MR 2188173 ( Abstract available ) LP1967a L.J. Lander and T.R. Parkin, “Consecutive primes in arithmetic progression,” Math. Comp., 21 (1967) 489. Rose94 H.E. Rose, A course in number theory, second edition, Clarendon Press, New York, 1994.

pp. xvi+398, ISBN 0-19-853479-5; 0-19-852376-9. MR 96g:11001 ( Annotation available ) Printed from the PrimePages © Reginald McLean.

## What is the nth term of the sequence 2 5 10 17 26?

Answer: The next three terms of the series 2, 5, 10, 17, 26,. are 37, 50, and 65 – Let’s solve this step by step. Explanation: The logic behind the terms of the given series is simple, the first term is 2 and the second term is 5 2 + 3 = 5 The third term is 10 5 + 5 = 10 The fourth term is 17 10 + 7 = 17 The fifth term is 26 17 + 9 = 26 Hence, we are adding consecutive odd numbers starting from 3 to each preceding term.

### What is the rule for 2 4 8 16?

Example: – 1, 2, 4, 8, 16, 32, 64, 128, 256,, This sequence has a factor of 2 between each number. Each term (except the first term) is found by multiplying the previous term by 2, In General we write a Geometric Sequence like this: where:

a is the first term, and r is the factor between the terms (called the “common ratio” )

### What is the sequence 1 2 4 6 8 16?

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

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### What is the nth term of the sequence 1 1 2 3 5 8?

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, The sequence you provided is a Fibonacci sequence. In a Fibonacci sequence, each number is the sum of the two preceding ones, starting with 1 and 1.

## What is the nth term rule for 1 2 5 8?

Solution: A sequence where the difference between each successive pair of terms is the same is called an Arithmetic sequence, The general rule of arithmetic sequence is a n = a 1 + (n – 1)d Where a 1 – first term d – common difference n – number of terms It is given that a 1 = -1 d = 2 + 1 = 3 Substituting it in the formula a n = -1 + (n – 1) 3 Therefore, the equation for the nth term is a n = -1 + (n – 1) 3.

#### What is the nth term of 9 11 13?

Summary: An equation for the nth term of the arithmetic sequence 9, 11, 13, 15, is a n = 2n + 7.

## What is the nth term of 7 11 15?

In the last section we learnt that we could use a formula which contains n to generate a sequence. To do this we set n as the position in the sequence. For the first term we set n = 1 and for the second term we set n = 2 etc. For the sequences 5 n, 5 n + 4 and 5 n – 3 we get the following results:

 5 n : 5, 10, 15, 20, 25,, (the five times table) 5 n + 4: 9, 14, 19, 24, 29,, 5 n – 3: 2, 7, 12, 17, 22,,

We can see that in each case the sequence goes up by 5 each time, and it turns out that any sequence which goes up by 5 each time will contain 5 n in the formula for the n th term of sequence. If the terms in a sequence go up by the same amount each time, then that amount will appear multiplied by n in the formula for the n th term of that sequence.e.g.

If the terms in a sequence go up by 7 each time, then 7 n will appear in the formula. Example: Finding a general formula for a sequence Imagine we want to find a formula for the n th term of this sequence: 7, 11, 15, 19, 23,, We can see that the terms in this sequence go up by 4 each time, so 4 n must appear in the formula,

The sequence generated by the formula 4 n is the four times table, but it isn’t quite the sequence we want:

 4 n sequence: 4, 8, 12, 16, 20,, The original sequence we wanted: 7, 11, 15, 19, 23,,

To get the terms in the original sequence, we need to add on 3 to each of the terms in the 4 n sequence. Hence, the general formula for the n th term of the original sequence is 4 n + 3.

### What is the nth term of 15 13 11 9?

The formula for the nth term of the sequence 15, 13, 11, 9, is 15 − 2 ( n − 1 ).

## What is the nth term for 1 3 6 10?

Answer and Explanation: 1, 3, 6, 10,. Therefore, the n t h term = n ( n + 1 ) 2.

### What is the nth term of 5 7 9 11?

Formula for the Nth Term of an Arithmetic Sequence – The formula to find the nth term of an arithmetic sequence is a n = a 1 + (n-1)d, where a n is the nth term, a 1 is the 1st term, n is the term number and d is the common difference. To find the formula for the nth term, we need a 1 and d.

1. To find the formula for the nth term of an arithmetic sequence, we need to know the difference, ‘d’, and the first term, ‘a 1 ‘.
2. In this example, the first term is 5 and the common difference is 2.
3. Substituting the values of a 1 = 5 and d = 2 into a n = a 1 + (n-1)d, we get a n = 5 + (n-1)2.
4. We simplify this by first expanding the brackets by multiplying (n-1) by 2.

We get a n = 5 + 2n – 2. This simplifies to a n = 2n + 3. : How to Find the Nth Term of an Arithmetic Sequence

## What is the nth term of 2 5 8 11 14?

The nth terms: 2,5,8,11,14, 17,20,23

## What is the nth term rule of 4 7 10 13?

The nth term for the sequence 4, 7, 10, 13 is n+3. if n= term number then 3n+1 will describe 4,7,10,13