can you help me find a method for the nth term of the sequence 9000, 8811, 8624, 8439, 8256?

Question

Can you help me to find a method for the nth term of the sequence 9000, 8811, 8624, 8439, 8256

Answer 1

nth term = n^2 - 192n + 9191

1st differences :
9000 - 8811 = 189
8811 - 8624 = 187
8624 - 8439 = 185
8439 - 8256 = 183

2nd differences :
189 - 187 = 2
187 - 185 = 2
185 - 183 = 2

3rd differences are all zero,
so equation to sequence is a quadratic,
of the form : an^2 + bn + c for n = 1,2,3, ...

n = 1 → a*1^2 + b*1 + c = 9000
n = 2 → a*2^2 + b*2 + c = 8811
n = 3 → a*3^2 + b*3 + c = 8624

or :

a + b + c = 9000
4a + 2b + c = 8811
9a + 3b + c = 8624

Three equations in three unknowns
can then be solved for a, b and c,
which results in a = 1, b = -192, c = 9191.

Answer 2

When ever the Diffence between copnsecutive terms of a series is consecutive odd number sthat mean that the nth term is a quadratic expression.

In case the difference of the consecutive terms is multiples of odd numbers, say 6, 10, 14 i.e the diference between the terms is 3X2, 5X2, 7X2, then 2 becomes the coeficient of the X^z term.

Now in this case,
The difference bet ween terms is 189, 187, 185, 183, that means that the series is an quadratic expression with the coefficient of X^2 being 1.

so, now the nth term here should be of the form. n^2 + p*n + q.
Now for n =1, the value is given as 9000,
that means 1 + p + q = 9000
and for n = 2; 4 +2 p +q = 8811;

solving these to equations u get the values of p and q,

Making the nth term:
n^2 - 192 n + 9191;


Enjoy!!!

Answer 3

9000-8811=189
8811-8624=187
8624-8439=185
You should have noticed that the difference is changing by 2 every time
Therefore the difference follows an arithmetic progression with common difference 2 and starting value -189

Since difference exist between two valus and not one,for the first term,i.e 9000 a difference cannot exist. between the first and second terms there is a difference
therefore the equation for the nth term for the DIFFERENCE is
D= -189+2(0.5n+1)

for an AP Nth term= a+(n-1)D
Nth term= 9000+[ -189+2(0.5n+1)][n-1]
=9000+(n*2 -192n+191)
=n*2 - 192n + 9191

Answer 4

1st differences
189,187,185,183
2nd differences
2,2,2
let N=nth term
N is a polynomial of second order
N=ax^2+bx+c
now use the terms
n=3,4,5
8624=9a+3b+c
8439=16a+4b+c
8256=25a+5b+c

sweep
-185=7a+b
-183=9a+b
-2=-2a>>>>a=1
b= -192,
8624=9-3*192+c
>>>c=9191

therefore, for n=1,2,3......
N=n^2-192n+9191

i hope that this helps

Answer 5

nth term = n^2 - 192*n + 9191

Answer 6

yes can you please tell me actual problem

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