Question

Find the nth term for the following sequences, 100,96,90,82,72.(an^2+bn+c)​

Answer 1

9514 1404 393

Answer:

  t(n) = -n^2 -n +102

Step-by-step explanation:

We can use the first 3 terms to find a, b, c.

  a(1)^2 +b(1) +c = 100

  a(2)^2 +b(2) +c = 96

  a(3)^2 +b(3) +c = 90

Subtract the first equation from the other two:

  3a +b = -4 . . . . . . [eq4]

  8a +2b = -10 . . . . [eq5]

[eq5] can be reduced to

  4a +b = -5 . . . . . . [eq6]

Subtracting [eq4] from [eq6] gives ...

  (4a +b) -(3a +b) = (-5) -(-4)

  a = -1 . . . . . . . . simplify

Filling that into [eq4], we get

  3(-1) +b = -4

  b = -1 . . . . . . . add 3

Putting the values for 'a' and 'b' into the first equation gives ...

  -1 +-1 +c = 100

  c = 102

Then the n-th term t(n) of the sequence is ...

  t(n) = -n^2 -n +102

_____

Additional comment

As you might guess, there is a generic solution for quadratic sequences. In terms of the first term (a1), first first difference (d1), and second difference (d2), the coefficients can be written as ...

• a = d2/2
• b = d1 -3a
• c = a1 -d1 +d2

This sequence has first differences of -4, -6, -8, -10, and second differences of -2. Putting the sequence values into the above formulas, we find ...

• a = -2/2 = -1
• b = -4 -3(-1) = -1
• c = 100 -(-4) +(-2) = 102 . . . . . as above