The first five terms of a sequence are 6, 6, 5, 3, 0.What is an expression for the nth term

Question

The value of n for whichthe nth term is -225

Answer 1

Series: 6 6 5 3 0

1st diff: 0 -1 -2 -3

2nd diff: -1 -1 -1

The 2nd difference is constant so the degree of the polynomial is 2, so the nth term, a[n]=an^2+bn+c where a, b, c are constants.

Let n=0 refer to the first term, so a0=6, and therefore c=6.

When n=1, a1=6=a+b+6 so a+b=0 and a=-b.

When n=2, a2=5=4a+2b+6=4a-2a+6=2a+6 and 2a=-1, a=-1/2, b=1/2.

a[n]=-n^2/2+n/2+6=n(1-n)/2+6. If n starts at 1 then we replace n by n-1: (n-1)(2-n)/2+6 or 6-(n-1)(n-2)/2.

If a[n]=-225, 6-(n-1)(n-2)/2=-225.

12-(n^2-3n+2)=-450, n^2-3n+2-450-12=0=n^2-3n-460=(n-23)(n+20), so n=23.