Given: 156; 148; 140; 132;...

Question

1) Determine the general formula for the nth term. 2) which term of this linear pattern is the first term to be negative. 3) the number pattern forms first differences of quadratic pattern Tn= an^2 +bn+c with t5=-24. Determine the general term of n

Answer 1

(1) This is an arithmetic progression (AP) with a common difference of -8.

The first term is 156 (when n=1)=156-8(n-1)=156-8n+8=164-8n. Formula is 164-8n.

(2) First term to become negative is when 164-8n<0, 164<8n, 41<2n, n>41/2. So n>20.5. The first integer is n=21. 164-8×21=164-168=-4.

The first negative term is the 21st term (=-4).

(3) T(n)=an²+bn+c is the quadratic formula such that:

Assuming n starts at 1, T(2)-T(1)=156, the first term of the AP.

T(3)-T(2)=148, T(4)-T(3)=140, T(5)-T(4)=132. 

T(5)=-24, so -24-T(4)=132, T(4)=-156.

So T(3)=T(4)-140=-296.

T(2)=T(3)-148=-444.

T(1)=T(2)-156=-600.

We need 3 equations to find 3 unknowns a, b, c.

(1) n=1: a+b+c=T(1)=-600;

(2) n=2: 4a+2b+c=T(2)=-444;

(3) n=3: 9a+3b+c=T(3)=-296.

(4)=(2)-(1): 3a+b=156,

(5)=(3)-(1): 8a+2b=304, 4a+b=152,

(6)=(5)-(4): a=-4, so b=152+16=168.

c=-600-a-b=-600+4-168=-764.

T(n)=-4n²+168n-764.

CHECK according to the formula for T(n):

T(1), T(2), T(3), T(4), T(5)=-600,-444,-296,-156,-24.

First differences: 156, 148, 140, 132 which is the given AP.