quadratic sequence are shown below Work out the next term.
7 11 17 25
2. The first four terms of a quadratic sequence are shown below Work out the next term.
6 12 22 36
3. The nth term of a quadratic sequence is n2 − 2n + 8 Work out the first three terms of this sequence

4. A quadratic sequence has an nth term of 2n2 + 3n − 1 Work out the value of the 6th term of the sequence


5. Asequencehasannthtermof n2−6n+7
Work out which term in the sequence has a value of 23.
6. Here are the first 5 terms of a quadratic sequence
4 11 20 31 44


Find an expression, in terms of n, for the nth term of this quadratic sequence.


7. Here are the first 5 terms of a quadratic sequence 4 10 18 28 40
Find an expression, in terms of n, for the nth term of this quadratic sequence.


8. Here are the first 5 terms of a quadratic sequence 9 17 29 45 65
Find an expression, in terms of n, for the nth term of this quadratic sequence.

9. Here is a tile.
Here is a sequence of patterns made from these tiles.
How many of these tiles are needed to make Pattern number 10?


10. The nth term of a sequence is n2 + 3n
Two consecutive terms in the sequence have a difference of 38
Work out the two terms.
11. Prove that every term in the sequence n2 − 4n + 21 is positive please help anyone

The quadratic sequence answered  this way .

Quadratic sequences are sequences that include an term. They can be identified by the fact that the differences between the terms are not equal, but the second differences between terms are equal.

According to the question

Quadratic sequences are given the next term is :

1> 7 11 17 25  

First difference between terms

11 - 7 = 4

17 - 11 = 6

25 - 17 = 8  

Second difference between terms

6 - 4 = 2

8 - 6 = 2

Therefore, next term : First last difference + Second difference  + Last Term

                                   : 8+2 = 10

                                   : 25 + 10 = 35

2> 6 12 22 36

First difference between terms

12 - 6 = 6

22 - 12 = 10

36 - 22 = 14

Second difference between terms

10 - 6 = 4

14 - 10 = 4

Therefore, 1st next term : First last difference + Second difference  + Last Term

                                     :  14 + 4 = 18

                                     :  36+ 18 = 54

2nd next term : First last difference + Second difference  + Last Term

                                     :  18 + 4 = 22

                                     :  22 + 54 = 76

3rd next term : First last difference + Second difference  + Last Term

                                     :   22 + 4 = 26

                                     :   76 + 26 = 102

4th next term : First last difference + Second difference  + Last Term

                                    :    26 + 4 = 30

                                     :   76 + 26 = 102

3> nth term of a quadratic sequence = [tex]n^{2} - 2n + 8[/tex]

first three terms of this sequence :

1st term of the sequence : n = 1

substituting value of n

[tex]1^{2} - 2 + 8[/tex]

= 7

2nd term of the sequence : n = 2

substituting value of n

[tex]2^{2} - 2*2 + 8[/tex]

= 8

3rd term of the sequence : n = 3

substituting value of n

[tex]3^{2} - 2*3 + 8[/tex]

= 17 - 6

= 11

4> Quadratic sequence has an nth term :  [tex]2n^{2} + 3n -1[/tex]

6th term of the sequence : n = 6

substituting value of n

[tex]2(6)^{2} + 3*6 -1[/tex]

36*2 + 18 -1

= 89

5> Quadratic sequence has an nth term : [tex]n^{2} - 6n + 7[/tex]

term = 23

therefore,

23 = [tex]n^{2} - 6n + 7[/tex]

0 = [tex]n^{2} - 6n - 16[/tex]

solving the equation

[tex]n^{2} - (8-2)n - 16 = 0[/tex]

n(n- 8) + 2 (n - 8) = 0

(n-8) (n+2) = 0

n = 8 , -2

As term can not be negative , so n = 8

Therefore, term 8 of Quadratic sequence  : [tex]n^{2} - 6n + 7[/tex] is 23 .

6> First 5 terms of a quadratic sequence :

4 11 20 31 44

expression for the nth term of this quadratic sequence:

As general form of quadratic sequence is : [tex]an^{2}+bn +c[/tex]

now for n= 1

a + b + c = 4 -------------------(1)

now for n= 2

[tex]a2^{2}+b*2 +c[/tex] = 11

4a + 2b + c = 11 ---------------------------(2)

now for n= 3

[tex]a3^{2}+b*3 +c[/tex] = 20

9a + 3b + c = 20  ----------------------------------(3)

solving equation (1) , (2) & (3)

a = 1 , b = 4 , c = -1

now , substituting the value in general equation

[tex]= n^{2}+4n -1[/tex]

Hence, The nth term of this quadratic sequence [tex]= n^{2}+4n -1[/tex]

7>The first 5 terms of a quadratic sequence : 4 10 18 28 40

the nth term of this quadratic sequence :

As general form of quadratic sequence is : [tex]an^{2}+bn +c[/tex]

now for n= 1

a + b + c = 4 -------------------(1)

now for n= 2

[tex]a2^{2}+b*2 +c[/tex] = 10

4a + 2b + c = 10 ---------------------------(2)

now for n= 3

[tex]a3^{2}+b*3 +c[/tex] = 18

9a + 3b + c = 18  ----------------------------------(3)

solving equation (1) , (2) & (3)

a = -4 , b = 18 , c = -10

now , substituting the value in general equation

[tex]= -4n^{2}+18n -10[/tex]

Hence, The nth term of this quadratic sequence [tex]= -4n^{2}+18n -10[/tex]

8>First 5 terms of a quadratic sequence : 9 17 29 45 65

the nth term of this quadratic sequence :

As general form of quadratic sequence is : [tex]an^{2}+bn +c[/tex]

now for n= 1

a + b + c = 9 -------------------(1)

now for n= 2

[tex]a2^{2}+b*2 +c[/tex] = 17

4a + 2b + c = 17 ---------------------------(2)

now for n= 3

[tex]a3^{2}+b*3 +c[/tex] = 29

9a + 3b + c = 29  ----------------------------------(3)

solving equation (1) , (2) & (3)

a = 2 , b = 2 , c = 5

now , substituting the value in general equation

[tex]= 2n^{2}+2n +5[/tex]

Hence, The nth term of this quadratic sequence [tex]= 2n^{2}+2n +5[/tex]

9> Pattern is not given (incomplete question)

10> The nth term of a sequence is : [tex]n^{2} + 3n[/tex]

Two consecutive terms in the sequence have a difference of 38

The two terms of the sequence  :

N2 - N1 = 38

substituting n as n2 and n1

[tex](N_{2}) ^{2} + 3N_{2} - (N_{1}) ^{2} - 3N_{1}[/tex] = 38

[tex](N_{2}) ^{2} - (N_{1}) ^{2} + 3N_{2} - 3N_{1} = 38[/tex]

[tex](N_{2} - N_{1} ) (N_{2} + N_{1})+ 3(N_{2} - N_{1}) = 38[/tex]

[tex](N_{2} - N_{1} ) [N_{2} + N_{1}+ 3] = 1*38\\\ (as (N_{2} - N_{1} = 1 ) consecutive terms)[/tex]

now,

[tex](N_{2} - N_{1} ) = 1 ,\\\ [N_{2} + N_{1}+ 3] = 38[/tex]

solving both equations

[tex]N_{2} =18\\ N_{1} = 17[/tex]

Hence, two terms of sequence are 18th and 17th the whose difference in 38 .

11> The sequence:  [tex]n^{2} - 4n + 21[/tex]

To proof  the sequence is positive always

As n cannot be negative

let n = 0

substituting the value in equation we get

+ve

As we conclude in equation +ve part is larger than -ve part always ,

so this equation is always be +ve for any value of n .

Hence, The quadratic sequence answered  this way .

To know more about quadratic sequence  here:

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