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Sagot :
1) 7, 11, 17, 25, 35
2) 6, 12, 22, 36, 54
6) The nth term is T= n^2 +4n -1
7) The nth term is T= n^2 + 3n
8) The nth term is T= 2n^2 + 2n + 5
I‘ll try to do the others, I’ll write the solutions in the comments of my answer
2) 6, 12, 22, 36, 54
6) The nth term is T= n^2 +4n -1
7) The nth term is T= n^2 + 3n
8) The nth term is T= 2n^2 + 2n + 5
I‘ll try to do the others, I’ll write the solutions in the comments of my answer
The quadratic sequence answered this way .
What are quadratic sequence?
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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