Question

what is the nth term rule of the quadratic sequence below /-7, -4, 3, 14, 29, 48, 71

Answer 1

Answer:

nth term = 2n^2 - 3n - 6.

Step-by-step explanation:

                      n   =  1   2    3    4     5     6    7  

                             - 7   -4   3   14   29   48   71  

Differences                3    7   11    15   19    23

Differences                   4   4     4    4     4

So the first term in the Rule is 2n^2

Subtracting  the values of 2n^2 from the original terms:

                  - 7   -4   3      14   29   48   71  

2n^2           2     8  18    32   50   72   98

Subtrct       -9   -12 -15  -18   -21   -24   -27

                       -3     -3    -3   -3    -3     -3

So we have -3n  and the last  term is  -6 because    -7 = 2(1)^2 -3 - 6  and the other terms fit this pattern as well.

Answer 2

The nth term rule of the provided quadratic sequence to find the n'th term of the sequence is 2n²-3n-6.

What is quadratic sequence?

The quadratic sequence is the series of the numbers in which the rule for the n'th term included an n squared term (n²). The quadratic equation for n can be given as,

$an^2+bn+c$

Here, (2a) is equal to the second difference, (3a+b) is equal to the first difference and (a+b+c) is equal to the first term.

The sequence given in the problem is

$-7, -4, 3, 14, 29, 48, 71$

Thus, (a+b+c) is,

$a+b+c=-7$                ......1

The first difference between the sequence is,

$3,7,11,15,19,23$

Thus, the value of (3a+b) is,

$3a+b=3\\b=3-3a$                      ......2

The second difference between the sequence is,

$4,4,4,4,4$

Thus, the value of 2a is,

$2a=4\\a=\dfrac{4}{2}=2$

Put this value in equation 2 as,

$b=3-3\times2\\b=-3$

Put the value of a and b in equation 1 as,

$2-3+c=-7\\c=-7+1\\c=-6$

Put these values in the above formula as,

$2n^2-3n-6$

The nth term rule of the provided quadratic sequence to find the n'th term of the sequence is,

$2n^2-3n-6$

Learn more about the quadratic sequence here;

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